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MATH 110 Sec 13.2 Operations with Events Practice Exercises If the probability of surviving a head-on car accident at 55 mph is 0.067, what is the probability of not surviving? MATH 110 Sec 13.2 Operations with Events Practice Exercises If the probability of surviving a head-on car accident at 55 mph is 0.067, what is the probability of not surviving? PROBABILITY OF THE COMPLEMENT OF AN EVENT If πΈ is an event, then π πΈ = 1 β π(πΈ β² ). MATH 110 Sec 13.2 Operations with Events Practice Exercises If the probability of surviving a head-on car accident at 55 mph is 0.067, what is the probability of not surviving? PROBABILITY OF THE COMPLEMENT OF AN EVENT If πΈ is an event, then π πΈ = 1 β π(πΈ β² ). π πππ‘ π π’ππ£ππ£πππ = 1 β π(π π’ππ£ππ£πππ) MATH 110 Sec 13.2 Operations with Events Practice Exercises If the probability of surviving a head-on car accident at 55 mph is 0.067, what is the probability of not surviving? PROBABILITY OF THE COMPLEMENT OF AN EVENT If πΈ is an event, then π πΈ = 1 β π(πΈ β² ). π πππ‘ π π’ππ£ππ£πππ = 1 β π(π π’ππ£ππ£πππ) π πππ‘ π π’ππ£ππ£πππ = 1 β 0.067 MATH 110 Sec 13.2 Operations with Events Practice Exercises If the probability of surviving a head-on car accident at 55 mph is 0.067, what is the probability of not surviving? PROBABILITY OF THE COMPLEMENT OF AN EVENT If πΈ is an event, then π πΈ = 1 β π(πΈ β² ). π πππ‘ π π’ππ£ππ£πππ = 1 β π(π π’ππ£ππ£πππ) π πππ‘ π π’ππ£ππ£πππ = 1 β 0.067 π πππ‘ π π’ππ£ππ£πππ = 0.933 MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: What is the probability of winning a free meal? MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 What is the probability that you do not win a free meal? MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 What is the probability that you do not win a free meal? COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 What is the probability that you do not win a free meal? COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) π πππβ² π‘ π€ππ = 1 β π π€ππ ππππ ππππ = 1 1 β 10 = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 What is the probability that you do not win a free meal? COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) π πππβ² π‘ π€ππ = 1 β π π€ππ ππππ ππππ = 1 1 β 10 = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises A cafe has scratch-and-win tickets with a chance of winning a free meal. If there is a 1 in 10 chance of winning a free meal: Saying that there is a 1 in 10 chance is just another way to say that the 1 probability of winning a free meal is . 10 What is the probability of winning a free meal? 1 π π€ππ ππππ ππππ = 10 What is the probability that you do not win a free meal? COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) π πππβ² π‘ π€ππ = 1 β π π€ππ ππππ ππππ = 1 1 β 10 = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: 0 1 2 Possible Number of Heads 3 4 5 6 7 8 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: 0 1 2 Possible Number of Heads 3 4 5 6 7 8 9 10 βAt least one headβ means one of these 9 possibilities MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: 0 1 2 Possible Number of Heads 3 4 5 6 7 8 9 10 βAt least one headβ means one of these 9 possibilities This makes it clear that it is MUCH easier to find π(0 βππππ ) and then use the Complement Rule. MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: 0 1 2 Possible Number of Heads 3 4 5 6 7 8 9 10 βAt least one headβ means one of these 9 possibilities This makes it clear that it is MUCH easier to find π(0 βππππ ) and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 9 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. 1 1023 9 = π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 1024 1024 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested the number heads. so visualize this: So, in letβs calculate π 0ofβππππ Because on eachPossible flip, we Number get eitherofHHeads or T (2 possibilities), Principle there are 0 the 1 Fundamental 2 3 Counting 4 5 6 says7that 8 9 10 10 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 = 1024 different that 10 coins canofbethese flipped. βAt least oneways headβ means one 9 possibilities BUT only 1 of these 1024 ways (TTTTTTTTTT) is β0 headsβ. 1 This makes it clear that MUCH=easier So, πit0isβππππ . to find π(0 βππππ ) 1024 and then use the Complement Rule. 1 1023 9 = π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 1024 1024 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises If 10 coins are flipped, what is the probability of obtaining at least one head? We are interested in the number of heads so visualize this: 0 1 2 Possible Number of Heads 3 4 5 6 7 8 9 10 This makes it clear that it is MUCH easier to find π(0 βππππ ) and then use the Complement Rule. 1 1023 9 = π ππ‘ ππππ π‘ πππ βπππ = 1 β π 0 βππππ = 1024 1024 10 MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? MATH 110 SecStandard 13-2 Lecture: Operations With Events deck of cards 4 suits (CLUBS, SPADES, HEARTS, DIAMONDS) 4 Queens If we select a single card from a standard 52-card deck, what is the probability that we draw either a heart or a face card? Let H be the event βdraw a heartβ and F be βdraw a face cardβ. That means that we are looking for π(π» βͺ πΉ). 13 HEARTS Recall what a standard 52-card deck looks like. MATH 110 SecStandard 13-2 Lecture: Operations With Events deck of cards 4 suits (CLUBS, SPADES, HEARTS, DIAMONDS) 4 Queens If we select a single card from a standard 52-card deck, what is the probability that we draw either a heart or a face card? Let H be theThere eventare βdraw a heartβ F be βdraw a face cardβ. 13 Hearts and 4and Queens BUT thethat Queen Hearts gets counted That means weofare looking for π(π»twice. βͺ πΉ). So, the number of Hearts or Queens is 13 + 4 β 1 = 16. 13 HEARTS Recall what a standard 52-card deck looks like. MATH 110 SecStandard 13-2 Lecture: Operations With Events deck of cards 4 suits (CLUBS, SPADES, HEARTS, DIAMONDS) 4 Queens If we select a single card from a standard 52-card deck, what is the probability that we draw either a heart or a face card? Let H be theThere eventare βdraw a heartβ F be βdraw a face cardβ. 13 Hearts and 4and Queens BUT thethat Queen Hearts gets counted That means weofare looking for π(π»twice. βͺ πΉ). So, the number of Hearts or Queens is 13 + 4 β 1 = 16. 13 HEARTS Recall what a standard 52-card deck looks like. MATH 110 SecStandard 13-2 Lecture: Operations With Events deck of cards 4 suits (CLUBS, SPADES, HEARTS, DIAMONDS) 4 Queens If we select a single card from a standard 52-card deck, what is the probability that we draw either a heart or a face card? Let H be theThere eventare βdraw a heartβ F be βdraw a face cardβ. 13 Hearts and 4and Queens BUT thethat Queen Hearts gets counted That means weofare looking for π(π»twice. βͺ πΉ). So, the number of Hearts or Queens is 13 + 4 β 1 = 16. 13 HEARTS Recall what a standard 52-card deck looks like. MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? There are 13 Hearts and 4 Queens BUT the Queen of Hearts gets counted twice. So, the number of Hearts or Queens is 13 + 4 β 1 = 16. MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? There are 13 Hearts and 4 Queens BUT the Queen of Hearts gets counted twice. So, the number of Hearts or Queens is 13 + 4 β 1 = 16. By the Basic Probability Principle # ππ π»ππππ‘π ππ ππ’ππππ 16 4 π π»ππππ‘ ππ ππ’πππ = = = # ππ πππππ 52 13 MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? There are 13 Hearts and 4 Queens BUT the Queen of Hearts gets counted twice. So, the number of Hearts or Queens is 13 + 4 β 1 = 16. By the Basic Probability Principle # ππ π»ππππ‘π ππ ππ’ππππ 16 4 π π»ππππ‘ ππ ππ’πππ = = = # ππ πππππ 52 13 MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? There are 13 Hearts and 4 Queens BUT the Queen of Hearts gets counted twice. So, the number of Hearts or Queens is 13 + 4 β 1 = 16. By the Basic Probability Principle # ππ π»ππππ‘π ππ ππ’ππππ 16 4 π π»ππππ‘ ππ ππ’πππ = = = # ππ πππππ 52 13 MATH 110 Sec 13.2 Operations with Events Practice Exercises What is the probability of getting either a heart or a queen when drawing a single card from a deck of 52 cards? There are 13 Hearts and 4 Queens BUT the Queen of Hearts gets counted twice. So, the number of Hearts or Queens is 13 + 4 β 1 = 16. By the Basic Probability Principle # ππ π»ππππ‘π ππ ππ’ππππ 16 4 π π»ππππ‘ ππ ππ’πππ = = = # ππ πππππ 52 13 MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) 0.7 = 0.3 + 0.65 β π(π΄ β© π΅) MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) 0.7 = 0.3 + 0.65 β π(π΄ β© π΅) MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) 0.7 = 0.3 + 0.65 β π(π΄ β© π΅) 0.7 = 0.95 β π(π΄ β© π΅) MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) 0.7 = 0.3 + 0.65 β π(π΄ β© π΅) 0.7 = 0.95 β π(π΄ β© π΅) π(π΄ β© π΅) = 0.95 β 0.70 MATH 110 Sec 13.2 Operations with Events Practice Exercises Assume A and B are events. If P(A U B) = 0.7, P(A) = 0.3 and P(B) = 0.65, find P(A β© B). UNION RULE FOR PROBABILITIES π π΄ βͺ π΅ = π π΄ + π π΅ β π(π΄ β© π΅) 0.7 = 0.3 + 0.65 β π(π΄ β© π΅) 0.7 = 0.95 β π(π΄ β© π΅) π(π΄ β© π΅) = 0.95 β 0.70 π(π΄ β© π΅) = 0.25 MATH 110 Sec 13.2 Operations with Events Practice Exercises The table below relates time consumers shop online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 MATH 110 Sec 13.2 Operations with Events Practice Exercises The table below relates time consumers shop online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals These are just subtotals. 121 453 182 552 270 547 573 1552 MATH 110 Sec 13.2 Operations with Events Practice Exercises The table below relates time consumers shop online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals These are just subtotals. 121 453 182 552 270 547 573 1552 This is the grand total (of consumers surveyed) MATH 13.2 Operations with Events Practice Exercises These are110 theSec numbers of consumers who shopped The table below relates time consumers shop online/month by income. 0-2the hrs/month online. Whatβs probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals These are just subtotals. 121 453 182 552 270 547 573 1552 This is the grand total (of consumers surveyed) MATH 13.2 Operations with Events Practice Exercises These are the numbers of These are110 theSec numbers of consumers with annual consumers who shopped The table below relates time consumers shop online/month by income. $60,000. 0-2the hrs/month online. Whatβs probability that a randomlyincomes selectedabove consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals These are just subtotals. 121 453 182 552 270 547 573 1552 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 182+171+121+156+270 = 900 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 182+171+121+156+270 = 900 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 182+171+121+156+270 = 900 0 β 2 βπ πππ ππ. ππ 900 π = =0.58 1552 ππππππ > $60000 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 182+171+121+156+270 = 900 0 β 2 βπ πππ ππ. ππ 900 π = =0.58 1552 ππππππ > $60000 This is the grand total (of consumers surveyed) MATH 110 Operations with Events Practice Because it is anSec OR,13.2 we are interested in the numbers thatExercises are inside of either or both rectangles The table below relates time consumers shopbelow. online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 270 547 573 1552 182+171+121+156+270 = 900 0 β 2 βπ πππ ππ. ππ 900 This is the grand total π = = 0.58 (of consumers surveyed) 1552 ππππππ > $60000 MATH 110 Sec 13.2 Operations with Events Practice Exercises The table below relates time consumers shop online/month by income. Whatβs the probability that a randomly selected consumer either spends 0-2 hrs/month shopping online or has an annual income above $60,000? (Round final Annual Income 10+ Hours 3-9 Hours 0-2 Hours Totals answer to 2 Above $60000 182 171 121 474 decimal $40000-60000 150 199 156 505 places.) Below $40000 Totals 121 453 182 552 182+171+121+156+270 = 900 0 β 2 βπ πππ ππ. ππ 900 π = = 0.58 1552 ππππππ > $60000 270 547 573 1552 MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? π Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 πππππ ππ ππππ π‘βππ = 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58 $1999 πππ₯π‘ ππππ‘β MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? π Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 πππππ ππ ππππ π‘βππ = 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58 $1999 πππ₯π‘ ππππ‘β MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? π Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 πππππ ππ ππππ π‘βππ = 0.06 + 0.13 + 0.22 + 0.34 + 0.13 =0.58 $1999 πππ₯π‘ ππππ‘β MATH 110 Sec 13.2 Operations with Events Practice Exercises An employee at an electronics store earns both a salary and a monthly commission as a sales rep. The table lists her estimates of the probability of earning various commissions next month. What is the probability that she will earn no more than $1,999 in commissions next month? π Commission Less than $1000 $1000-$1249 $1250-$1499 $1500-$1749 $1750-$1999 $2000-$2249 $2250-$2499 $2500 or more Probability 0.06 0.13 0.22 0.34 0.13 0.04 0.05 0.03 πππππ ππ ππππ π‘βππ = 0.06 + 0.13 + 0.22 + 0.34 + 0.13 = 0.58 $1999 πππ₯π‘ ππππ‘β MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M S 0.15 M is βneeds new motorβ. S is βneeds new switchβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S 0.15 =1 ? MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S 0.15 =1 0.15 MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S 0.15 0.15 MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S π 0.15 0.15 πππππ = ππππ‘βππ MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S π 0.15 0.15 πππππ = 0.15 ππππ‘βππ MATH 110 Sec 13.2 Operations with Events Practice Exercises The windshield wipers on a car have not been working properly. The probability that the car needs a new motor is 0.55, that it needs a new switch is 0.45 and the probability that it both is 0.15. What is the probability that the car needs neither a new motor nor a new switch? Venn diagrams can be useful in solving problems like this. M M is βneeds new motorβ. S is βneeds new switchβ. S π 0.15 0.15 πππππ = 0.15 ππππ‘βππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. P is βis on academic probationβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises In other words, there students are 62 students in circle P A study of 205 randomly selected examined the relationship but 22 academic of those are NOT inand circle S. between satisfaction with advising academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. P is βis on academic probationβ. MATH 110 Sec 13.2 Operations with Events Practice Exercises In other words, there students are 62 students in circle P A study of 205 randomly selected examined the relationship but 22 academic of those are NOT inand circle S. between satisfaction with advising academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. P is βis on academic probationβ. 22 MATH 110 Sec 13.2 Operations with Events Practice Exercises In other words, there students are 62 students in circle P A study of 205 randomly selected examined the relationship but 22 academic of those are NOT inand circle S. between satisfaction with advising academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. P is βis on academic probationβ. 22 MATH 110 Sec 13.2 Operations with Events Practice Exercises In other words, there students are 62 students in circle P A study of 205 randomly selected examined the relationship but 22 academic of those are NOT inand circle S. between satisfaction with advising academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. P is βis on academic probationβ. 22 But if there are 62 students in circle P, then there must be 40 students here. MATH 110 Sec 13.2 Operations with Events Practice Exercises In other words, there students are 62 students in circle P A study of 205 randomly selected examined the relationship but 22 academic of those are NOT inand circle S. between satisfaction with advising academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 P is βis on academic probationβ. 22 But if there are 62 students in circle P, then there must be 40 students here. MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 P is βis on academic probationβ. 22 MATH 110 Sec 13.2 Operations with Events Practice Exercises So, only 25 students not in examined the relationship A study of 205 randomly selected students circle P are also notadvising in S. and academic success. Of between satisfaction with academic the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 P is βis on academic probationβ. 22 MATH 110 Sec 13.2 Operations with Events Practice Exercises So, only 25 students not in examined the relationship A study of 205 randomly selected students circle P are also notadvising in S. and academic success. Of between satisfaction with academic the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 P is βis on academic probationβ. 22 25 MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P 40 22 There are 205 students in all and we have already accounted for 40+22+25 = 87 of them. 25 So, there must be 205-87 = 118 here. MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P 40 22 There are 205 students in all and we have already accounted for 40+22+25 = 87 of them. 25 So, there must be 205-87 = 118 here. MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P 118 40 22 There are 205 students in all and we have already accounted for 40+22+25 = 87 of them. 25 So, there must be 205-87 = 118 here. MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 118 P is βis on academic probationβ. 22 25 MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 118 P is βis on academic probationβ. 22 25 MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 118 P is βis on academic probationβ. 22 25 MATH 110 Sec 13.2 Operations with Events Practice Exercises A study of 205 randomly selected students examined the relationship between satisfaction with academic advising and academic success. Of the 62 students on academic probation, 22 are not satisfied with their advising. But, only 25 of the students not on academic probation are dissatisfied with their advising. Whatβs the probability that a randomly selected student is on academic probation and is satisfied with advising? Round final answer to 2 decimal places. S P S P S is βsatisfied with advisingβ. 40 118 P is βis on academic probationβ. 22 40 ππ ππππππ‘πππ = 0.20 25 π πππ π ππ‘ππ ππππ = 205 MATH 110 Sec 13.2 Operations with Events Practice Exercises A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? MATH 110 Sec 13.2 Operations with Events Practice Exercises A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. MATH 110 Sec 13.2 Operations with Events Practice Exercises A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) MATH 110 Sec 13.2 Operations with Events Practice Exercises A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. A company shipped 57 cameras to a store. Two of these are defective. If the store sold 20 cameras before discovering that some were defective, what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ Keystrokes: 5 557 205557 20 MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ Keystrokes: 5 557 205557 20 MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ ππ‘ ππππ π‘ 1 π = 1 β0.417 ππππππ‘ππ£π π πππ Keystrokes: 5 557 205557 20 MATH 110 SecBut 13.2the Operations with Events Practice hint told us how to calculate thisExercises probability. πΆ(55,20) A company shipped 57 cameras to a store. Two of these are defective. ππ ππππππ‘ππ£π π before discovering = that some were = 0.417 If the store sold 20 cameras defective, πΆ(57,20) π πππ what is the probability that at least one defective camera was sold? HINT: πͺ(ππ,ππ) πͺ(ππ,ππ) is the probability that no defective cameras were sold. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ 1 ππ ππππππ‘ππ£π π =1βπ ππππππ‘ππ£π π πππ π πππ ππ‘ ππππ π‘ 1 π = 1 β0.417 ππππππ‘ππ£π π πππ = 0.583 MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability that at least one customer will receive spoiled food? Give final answer as a simplified fraction. MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability that at least one customer will receive spoiled food? Give final answer as a simplified fraction. ππππ πππ‘ As in the previous problem, it is easier to find π first π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability that at least one customer will receive spoiled food? Give final answer as a simplified fraction. ππππ πππ‘ As in the previous problem, it is easier to find π first π ππππππ ππππ and then use the Complement Rule. MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability that at least one customer will receive spoiled food? Give final answer as a simplified fraction. ππππ πππ‘ As in the previous problem, it is easier to find π first π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability that at least one customer will receive spoiled food? Give final answer as a simplified fraction. ππππ πππ‘ As in the previous problem, it is easier to find π first π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘ As in the previous problem, it is easier to find π first π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs As in the previous problem, it is easier to find π first π ππππππ ππππ done here, but we will ignore that hint and just work it all out.) and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs As in the previous problem, it is easier to find π first π ππππππ ππππ done here, but we will ignore that hint and just work it all out.) and thenthe useBasic the Complement Rule. First apply Probability Principle COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs As in the previous problem, it is easier to find π first π ππππππ ππππ done here, but we will ignore that hint and just work it all out.) and thenthe useBasic the Complement Rule. First apply Probability Principle β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec Operations Practice Exercises To calculate the13.2 number of ways with to getEvents the portions, spoiled or not, we can choose ANY of the 23 2 portions. A sushi shop prepared 23 portions of seafood, of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec Operations Practice Exercises To calculate the13.2 number of ways with to getEvents the portions, spoiled or not, we can choose ANY of the 23 2 portions. A sushi shop prepared 23 portions of seafood, of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec Operations Practice Exercises To calculate the13.2 number of ways with to getEvents the portions, spoiled or not, we can choose ANY of the 23 2 portions. A sushi shop prepared 23 portions of seafood, of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability thatofatways leasttoone will receive food? To find the number get customer onlyππππ GOODπππ‘ portions , wespoiled can choose letβs find πasportions . were spoiled). only from among the 21 GOOD (because 2 GiveSo, final answer a simplified fraction. π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is the probability thatofatways leasttoone will receive food? To find the number get customer onlyππππ GOODπππ‘ portions , wespoiled can choose letβs find πasportions . were spoiled). only from among the 21 GOOD (because 2 GiveSo, final answer a simplified fraction. π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ ππππ πππ‘to be (The hint on the previous problem tells us exactly what needs In either case,problem, we are selecting 18 portions As in the previous it is easier to find but π obviously order first π ππππππ done here, butThis we tells will ignore hintcount and just work it allππππ out.) doesnβt matter. us thatthat we will using combinations. and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) MATH 110 Sec 13.2 Operations with Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) Letβs do the denominator first.Practice Exercises MATH 110 Sec 13.2 Operations with Events A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) Letβs do the denominator first.Practice Exercises MATH 110 Sec 13.2 Operations with Events A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served 33649 long and Keystrokes: spoiled. If 182of3the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) Letβs do the denominator first.Practice Exercises MATH 110 Sec 13.2 Operations with Events A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served 33649 long and Keystrokes: spoiled. If 182of3the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ πΆ(23,18) Letβs do the denominator first.Practice Exercises MATH 110 Sec 13.2 Operations with Events A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served 33649 long and Keystrokes: spoiled. If 182of3the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 Now do the with numerator. MATH 110 Sec 13.2 Operations Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too long and spoiled. If 18 of the 23 portions are served randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 Now do the with numerator. MATH 110 Sec 13.2 Operations Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served 33649 long and Keystrokes: spoiled. If 182of1the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 Now do the with numerator. MATH 110 Sec 13.2 Operations Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served1330 long and Keystrokes: spoiled. If 182of1the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18)= 1 β π π ππππππ ππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 Now do the with numerator. MATH 110 Sec 13.2 Operations Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served1330 long and Keystrokes: spoiled. If 182of1the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 Now do the with numerator. MATH 110 Sec 13.2 Operations Events Practice Exercises A sushi shop prepared 23 portions of seafood, 2 of which stayed out too 1 8 are served1330 long and Keystrokes: spoiled. If 182of1the 23 portions randomly, what is ππππ πππ‘ the probability that at least one customer will receive spoiled food? letβs find πas a simplified fraction. . GiveSo, final answer π ππππππ ππππ Because weon must the answer astells a simplified fraction (unliketoinbethe (The hint the give previous problem us exactly what needs previous problem), weignore will notthat be able to just done here, but we will hint and justpunch work the it allentire out.) expression into the calculator and get the final answer. and then use Rule. and reduce. Instead, we must calculator thethe topComplement and bottom separately β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ πΆ(21,18) ππππ 1330 = 1 β π π ππππππ ππππ π ππππππ ππππ π = π ππππππ ππππ 33649 we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ this strategy, see if 21 π πππ‘ = 1 βApplying π πΆ(21,18) ππππ 1330 π ππππππ ππππ π ππππππ divides both top and bottom π = ππππ π ππππππ ππππ 33649 evenly.divide 1330 Oops! 21 doesnβt evenly so move onβ¦trying 7 next. we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ this strategy, see if 21 π πππ‘ = 1 βApplying π πΆ(21,18) ππππ 1330 π ππππππ ππππ π ππππππ divides both top and bottom π = ππππ π ππππππ ππππ 33649 evenly.divide 1330 Oops! 21 doesnβt evenly so move onβ¦trying 7 next. we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 1330 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 33649 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 1330 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 33649 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ 7 × 190 π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 1330 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 33649 YES! 7 × 4807 (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ 7 × 190 π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 7 × 190 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 7 × 4807 YES! 7 × 4807 (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 7 × 190 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 7 × 4807 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 7 × 190 \ π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ \7 × 4807 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 β π Does 7 divide both πΆ(21,18) ππππ 190 π ππππππ ππππevenly? π ππππππ top and bottom π = ππππ π ππππππ ππππ 4807 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 βNow π the top is small enough to πΆ(21,18) ππππ 190 π ππππππ π ππππππ easily factor intoππππ its prime factors. π = ππππ π ππππππ ππππ 4807 YES! (1330/7 = 190 & 33649/7 = 4807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 βNow π the top is small enough to πΆ(21,18) ππππ 190 π ππππππ π ππππππ easily factor intoππππ its prime factors. π = ππππ π ππππππ ππππ 4807 YES! 190 = 19 x 10 = 19 x 5 x 24807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1 βNow π the top is small enough to πΆ(21,18) ππππ 19 × 5 × 2 π ππππππ π ππππππ easily factor intoππππ its prime factors. π = ππππ π ππππππ ππππ 4807 YES! 190 = 19 x 10 = 19 x 5 x 24807) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β πtest each factor on top to see πΆ(21,18) ππππ 19 × 5 × 2 π ππππππ ππππ π ππππππ if it evenly divides the bottom. π = ππππ π ππππππ ππππ 4807 YES! ) we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β πtest each factor on top to see πΆ(21,18) ππππ 19 × 5 × 2 π ππππππ ππππ π ππππππ if it evenly divides the bottom. π = ππππ π ππππππ ππππ 4807 YES! 4807/19 = 253 so 19 works! we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β πtest each factor on top to see πΆ(21,18) ππππ 19 × 5 × 2 π ππππππ ππππ π ππππππ if it evenly divides the bottom. π = ππππ π ππππππ ππππ 19 × 253 YES! 4807/19 = 253 so 19 works! we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β πtest each factor on top to see πΆ(21,18) ππππ 19 × 5 × 2 π ππππππ ππππ π ππππππ if it evenly divides the bottom. π = ππππ π ππππππ ππππ 19 × 253 YES! 4807/19 = 253 so 19 works! we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β πtest each factor on top to see πΆ(21,18) ππππ 5 × 2 π ππππππ ππππ π ππππππ if it evenly divides the bottom. π = ππππ π ππππππ ππππ 253 YES! 4807/19 = 253 so 19 works! we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Finally, β π test 5 and 2 (on top) to see πΆ(21,18) ππππ 5 × 2 ππππ the bottom. π ππππππ if eitherπ ππππππ evenly divides π = ππππ π ππππππ ππππ 253 YES! we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Finally, β π test 5 and 2 (on top) to see πΆ(21,18) ππππ 5 × 2 ππππ the bottom. π ππππππ if eitherπ ππππππ evenly divides π = ππππ π ππππππ ππππ 253 Neither 5 nor 2 evenly divides 253 thus we canβt simplify further. we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Finally, β π test 5 and 2 (on top) to see πΆ(21,18) ππππ 10 ππππ the bottom. π ππππππ if eitherπ ππππππ evenly divides π = ππππ π ππππππ ππππ 253 Neither 5 nor 2 evenly divides 253 thus we canβt simplify further. we13.2 must reduce thewith fraction to lowest terms. MATHFinally, 110 Sec Operations Events Practice Exercises larger23 numbers this be time-consuming. A sushi shopWith prepared portions ofcan seafood, 2 of which stayed out too However, here strategy that might help. long and spoiled. If 18 of theis23a portions are served randomly, what is back atthat the at combination that produced the numeratorβ¦ ππππ πππ‘ theLook probability least one customer will receive spoiled food? So, letβs findtest πas to . in this case,Give πΆ(21,18) and see if the fraction will reduce final answer a simplified fraction. π ππππππ ππππ by the first numberβ¦21 in this case. If it works, great, if not, test (The the hintother on thefactors previous tells usfrom exactly what ofproblem the number small to needs largeβ¦to be done here, but we willthose ignorewould that hint just here be and 7 and 3.work it all out.) and then use the Complement Rule. β²) # ππ π€ππ¦π π‘π πππ‘ ππ π ππππππ ππππ πππ‘ COMPLEMENT RULE: π πΈ = 1 β π(πΈ π = π ππππππ ππππ # π€ππ¦π ππππ πππ‘ππ πππ‘) ππ‘ ππππ π‘ πππ πππ‘ π‘π πππ‘ πππ¦ (π ππππππ π πππ‘ = 1Now β π that we have this probability, πΆ(21,18) ππππ 10 π ππππππ π ππππππ we can go backππππ and plug it in to π = ππππ π ππππππ ππππ 253 get the final result that we were looking for in the beginning. MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππππ πππ‘ ππ‘ ππππ π‘ πππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ πππ πππ‘ 10ππππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ 253 MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) ππ‘ ππππ π‘ πππ πππ‘ 10ππππ πππ‘ π =1βπ π ππππππ ππππ π ππππππ ππππ 253 MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) πππ‘ 10 243 ππ‘ ππππ π‘ πππ πππ‘ 10ππππ253 π =1βπ = ππππ β = π ππππππ π ππππππ ππππ 253 253 253 253 MATH 110 Sec 13.2 Operations with Events Practice Exercises # ππ π€ππ¦π π‘π πππ‘2ππ π ππππππ ππππprepared πππ‘ A sushi shop 23 portions of seafood, of which stayed out too π = π ππππππ ππππ π€ππ¦π π‘π πππ‘ πππ¦ ππ πππ‘) what is long and spoiled. If 18 of#the 23 portions are(π ππππππ served randomly, Now that wereceive have this probability, the probability least will spoiled food? ππππ πππ‘that at πΆ(21,18) 10one customer we can go fraction. back and plug it in to π = answer as a simplified Give final π ππππππ ππππ 253 get the final result thatπππ‘ we were ππππ ππππ ππ‘ ππππ π‘ πππ πππ‘ looking for inπππ‘ the beginning. first As in the previous problem, it is easier to find π π =1βπ π ππππππ π ππππππ ππππ ππππ π ππππππ ππππ and then use the Complement Rule. COMPLEMENT RULE: π πΈ = 1 β π(πΈ β² ) πππ‘ 10 243 ππ‘ ππππ π‘ πππ πππ‘ 10ππππ253 π =1βπ = ππππ β = π ππππππ π ππππππ ππππ 253 253 253 253