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Probability of Independent and Dependent Events CCM2 Unit 6: Probability Independent and Dependent Events • Independent Events: when one event has no affect on the probability of the other event occurring. • Dependent Events: if the outcome or probability of the first event affects the outcome of the second. Independent Events Suppose a die is rolled and then a coin is tossed. • Explain why these events are independent. – They are independent because the outcome of rolling a die does not affect the outcome of tossing a coin, and vice versa. • Fill in the table to represent the sample space – how many total outcomes are there? Roll 1 Head Tail Roll 2 Roll 3 Roll 4 Roll 5 Roll 6 Roll 1 Roll 2 Roll 3 Roll 4 Roll 5 Roll 6 Head 1,H 2,H 3,H 4,H 5,H 6,H Tail 1,T 2,T 3,T 4,T 5,T 6,T • How can you find the total outcomes of the sample space WITHOUT writing it out? Remember the Fundamental Counting Principle. – Still stumped? Here is the process: – How many outcomes are there for rolling the die? • 6 outcomes – How many outcomes are there for tossing the coin? • 2 outcomes – How many outcomes are there in the sample space of rolling the die and tossing the coin? 6 • 2 = 12 outcomes Roll 1 Roll 2 Roll 3 Roll 4 Roll 5 Roll 6 Head 1,H 2,H 3,H 4,H 5,H 6,H Tail 1,T 2,T 3,T 4,T 5,T 6,T Use the table to find the following probabilities: 1. P(rolling a 3) 2/12 = 1/6 2. P(Tails) 6/12 = ½ 3. P(rolling a 3 AND getting tails) 1/12 What do you notice about #1 and #2 to get #3? 4. P(rolling an even) 6/12 = ½ 5. P(heads) 6/12 = ½ 6. P(rolling an even AND getting heads) 3/12 or 1/4 Do you see the pattern again with #4 and #5 to get #6? Multiplication Rule of Probability for Independent Events • The probability of two independent events occurring can be found by the following formula: P(A and B) = P(A) x P(B) • Let’s go back to the table • How can we find the probabilities for this table? – Just the way you found the total number of possibilities – by multiplying! Head P(H) = 1/2 Tail P(T) = 1/2 Roll 1 Roll 2 Roll 3 P(1) = 1/6 P(2) = 1/6 P(3) = 1/6 Roll 4 Roll 5 P(4) = 1/6 P(5)= 1/6 Roll 6 P(6)= 1/6 1,H 2,H 3,H 4,H 5,H 6,H P(1 and H) = P(2 and H) = P(3 and H) = P(4 and H) = P(5 and H) = P(6 and H) = (1/6)(1/2) = 1/12 1/12 1/12 1/12 1/12 1/12 1,T 2,T 3,T 4,T 5,T 6,T P(1 and T) = P(2 and T) = P(3 and T) = P(4 and T) = P(5 and T) = P(6 and T) = 1/12 1/12 1/12 1/12 1/12 1/12 Example 1 At City High School, 30% of students have parttime jobs and 25% of students are on the honor roll. What is the probability that a student chosen at random has a part-time job and is on the honor roll? Write your answer in context. P(PT job and honor roll) = P(PT job) x P(honor roll) = .30 x .25 = .075 There is a 7.5% probability that a student chosen at random will have a part-time job and be on the honor roll. Dependent Events • Remember, we said earlier that • Dependent Events: two events are dependent if the outcome or probability of the first event affects the outcome or probability of the second. • Let’s look at some scenarios and determine whether the events are independent or dependent. Example 2: Determine whether the events are independent or dependent: a.) Selecting a marble from a container and selecting a jack from a deck of cards. • Independent b.) Rolling a number less than 4 on a die and rolling a number that is even on a second die. • Independent c.) Choosing a jack from a deck of cards and choosing another jack, without replacement. • Dependent d.) Winning a hockey game and scoring a goal. • Dependent Multiplication Rule of Probability for Dependent Events • The probability of two dependent events occurring can be found by the following formula: P(A and B) = P(A) x P(B following A) Example 3: Let’s look at a different of Independent vs. Dependent Example INDEPENDENT A box contains 4 red marbles and 6 purple marbles. You are going to choose 3 marbles with replacement . a.) Draw a tree diagram and write the probabilities on each Purple branch. Purple Purple Red Red Purple Red Red Purple Red Purple Red Purple Red b.) What is the probability of drawing 2 purple marbles and 1 red marble in succession (aka in order)? P(1st draw purple) • P(2nd draw purple) • P(3rd draw red) 6/10 • 6/10 • 4/10 = 18/125 or .144 or 14.4% DEPENDENT A box contains 4 red marbles and 6 purple marbles. You are going to choose 3 marbles without replacement . a.) Draw a tree diagram and write the probabilities on each Purple branch. Purple Purple Red Red Purple Red Red Purple Red Purple Red Purple Red b.) What is the probability of drawing 2 purple marbles and 1 red marble in succession (aka in order)? P(1st draw purple) • P(2nd draw purple) • P(3rd draw red) 6/10 • 5/9 • 4/8 = 1/6 or .167 or 16.7% In the above example, what is the probability of first drawing all 4 red marbles in succession and then drawing all 6 purple marbles in succession without replacement? • P(4 red then 6 purple) = (4/10)(3/9)(2/8)(1/7)(6/6)(5/5)(4/4)(3/3)(2/2)(1/1) = 1/210 or .0048 • The probability of drawing 4 red then 6 purple without replacement is 0.48% • Explain why the last 6 probabilities above were all equivalent to 1. • This is because there were only purple marbles left, so the probability for drawing a purple marble was 1. Example 4: How does permutations and combinations work into this type of probability??? a.) There are 32 students in the classroom and the teacher needs to be make a seating chart. What is the probability that Deshawn will be in the first seat, Sophie in the second seat, and Delaney in the third seat? P(Deshawn 1st) • P(Sophie 2nd) • P(Delaney 3rd ) 1/32 • 1/31 • 1/30 = 3.36 x 10-5 = .0000336 = .00336% You get the same answer! OR Since the order matters, we call this a permutation choosing 3 out of 32: 32P3 and we’re looking for a specific one order, so: 1_ = 1 = .00336% 32•31•30 32P3 Example 4: How does permutations and combinations work into this type of probability??? b.) There are 32 students in the classroom and the teacher can only choose 2 students to go to the media center. What is the probability that Arturo and Jon will be chosen? P(Arturo or Jon) • P(the other one) 2/32 • 1/31 = .002 = .20% You get the same answer! OR Since the order does NOT matter, we call this a combination choosing 2 out of 32: 32C2 and we’re looking for a specific one group of people, so: 1_ = 1 = .20% 32 •31 32C2 2•1