Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MDM4U Unit 1 Master Overview of Unit 1: Introduction to Probability (Chapter # 4 – Nelson) Lesson 1: Lesson 2: Lesson 3: Lesson 4: Lesson 5: Lesson 6: Lesson 7: Lesson 8: Lesson 9: Lesson 10: Lesson 11: Lesson 12: Lesson 13: Sec.4.1 An Introduction to Simulations Sec.4.2 Theoretical Probability Sec.4.3a. Finding Probability Using Sets Sec.4.3b. Finding Probability Using Venn Diagrams Sec.4.4 Conditional Probability Sec.4.5 Finding Probability Using Tree Diagrams Sec.4.6a Permutations Sec.4.6b Probability & Permutations Sec.4.7a Combinations Sec.4.7b Probability & Odds Sec.4.7c Problem Solving with Combinations Unit Review Day # 1 Unit Review Day # 2 Unit 1: Homework Topic Homework Simulations p.210 # 4, 5, 8, 10, 11, 14, 19 Theoretical Probability p.218 # 1 , 2b , 3 – 6 , 8 , 10 , 11, 14 , 18 Probability Using Sets p.228 # 1abcdf , 6a , 8a Probability Using Venn Diagrams p.228 # 2 , 3 , 4 , 6b , 7 , 8b , 9 , 12 , 13 , 14 Conditional Probability p.236 # 1 , 2 , 5 , 8 , 9 , 14ab Tree Diagrams p.245 # 3 , 4ab , 5abc (ii,iii,iv) , 6ab , 9ab Permutations p.255 # 1bde , 2b , 3c , 4abc , 5 , 6ab , 9a , 11 , 13 , 15 - 17 Probability & Permutations YES Combinations p.262 # 1 - 3 , 6 , 7 omit g Odds YES Problem Solving with Combinations YES Page 1 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Probability & Odds Homework: 1. a) Given: P A 7 36 Find: odds in favour of A b) Given: P A ' 2 6 Find: odds in favour of A c) Given: (odds against A) = 5 : 4 d) Given: (odds in favour of A) = 10 : 8 (odds against B) = 5 : 1 A & B are mutually exclusive Find: (i) (ii) e) Find: P(A) P A B odds in favour of either A or B occurring Given: P(A) = 0.5 P(B) = 0.3 A & B are independent Find: odds in favour that both A & B occur [ 7:29 , 2:1 , 4/9 , 13/18 , 13:5 , 3:17 ] 2. h h Show that if (odds in favour of A) = then P A k hk 3. What are the odds in favour of rolling a pair, followed by two even numbers in two successive rolls of a pair of dice? 4. Over time, Natalie has experienced that if she washer her car, it rains the next day 80% of the time. What are the odds in favour of rain tomorrow if Natalie is washing her car today? [ 1:23 ] [ 4:1 ] 5. Statistics show that in 75% of the fatal accidents involving two cars, at least one of the drivers is impaired. If you have the misfortune to witness such an accident, what are the odds in favour of one of the drivers being impaired? [ 3:1 ] Page 2 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Probability & Permutations Homework: 1. 2. 3. Match each expression on the left with an equivalent expression on the right. 14 ! a) i) 10 100 13! 52 ! b) ii) 6! 51! 101! c) iii) 52 99! d) 20 19! iv) 10! e) 90 x 8! v) 14 f) 30 x 4! vi) 20! Find the value for each expression. a) 8! 5! b) 19 ! 13 ! e) 155 ! 152 ! f) 93! 89 ! 4 ! c) 21! 17 ! 4 ! 9! 7 ! 2! d) [ 336 , 19535040 , 5985 , 36 , 3652110 , 2919735 ] Evaluate each of the following expressions, using the definition of the factorial operation to simplify the work. a) 10 ! 5! b) 21! 14 ! c) 9! 3 ! 6! e) 7! 7! 2 ! 5! 4 !3! f) 15! 15! 9 ! 6! 10!5! g) 2 5! 3! 2 ! d) 12 ! 8 ! 4! h) 3 11! 7 !4! [ 30240 , 586051200 , 84 , 495 , 56 , 8008 , 20 , 990 ] 4. Simplify the following expressions a) nn 1! d) n! n 2 3n 2 b) n!n 1 c) n 1!n 2 n e) n! n 2! f) n 2! n 1! [ n! , (n+1)! , (n+1)! , (n+2)! , n2-n , (n+2)(n+1)n ] Page 3 of 45 Prepared by: Scott McEwen MDM4U 5. Unit 1 Master Solve for n, n W a) c) n 1! 9 n! 3n 1! 126 n 1! b) n! 20 n 2! d) 2n! 84n n 3! [8,5,6,8] 6. In how many ways can the letters of each word be arranged? a) d) MAXIMUM INTERESTING b) e) CANADA UNINTERESTING c) f) SASKATCHEWAN MISSISSAUGA [ 840 , 120 , 39916800 , 2494800 , 129729600 , 415800 ] 7. List all five-digit numbers than can be formed by using two 4’s and three 6’s. 8. How many seven-digit integers, are there which include a) two 3’s, three 2’s and two 8’s? b) four 3’s and three 4’s? [ 10 ] [ 210 , 35 ] 9. A man bought two vanilla ice cream cones, three chocolate cones, four strawberry cones, and one pistachio cone for his ten children. In how many ways can he distribute the flavours among the children? [ 12600 ] 10. A coin is tossed nine times. In how many ways could the results be six heads and three tails? 11. Anya is starting out on her evening run. Her route always takes her eight blocks east and five blocks north to her grandmother’s apartment building. But she likes to vary the path she follows. How many different possibilities does she have? Hint: Consider the permutations of 13 letters, 8E’s and 5 N’s 12. How many numbers greater than 300000 are there using only the digits 1 , 1 , 1 , 2 , 2 , 3 ? 13. Yurak is shelving books in a display in the school library. He has four different books with three copies of each. In how many ways can he arrange the books on the shelf for display? [ 84 ] [ 1287 ] [ 10 ] [ 369600 ] 14. A developer will build 12 houses on the same side of Costly Court in a new subdivision. If he has room for two houses modeled on Plan A, four modeled on Plan B, and six modeled on Plan C, in how many different ways can he arrange the houses on the street? [ 13860 ] 15. Emily’s minor soccer team played a total of 14 games in the season. Their record was eight wins, four losses, and two ties. In how many orders could this have happened? [ 45045 ] 16.a) b) c) d) How many permutations are there of the letters of the word BASKETBALL? How many of the arrangements begin with K? How many of the arrangements start with a B? In how many of the arrangements would the two L’s be together? [ 453600 , 45360 , 90720 , 90720 ] Page 4 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Problem Solving in Combinations Homework: 1. In how many ways can a committee of at least one person be formed from seven club members? 2. Chang arrives at the giant auction sale late in the afternoon. There are only five items left to be sold. How many different purchases could he make? 3. In how many different ways could a team of three students be chosen from Lin’s Finite Mathematics class of 25 students to complete in the Country Mathematics Contest? 4.a) How many different sums of money can be made from a $2-bill, a $5-bill, and a $10-bill? [ 127 ] [ 31 ] [ 2300 ] [7] b) How many different sums of money can be made from the bills in (a) as well as one more $10-bill? [ 11 ] c) Why does the situation become much more complicated if another $5-bill is added? 5. A committee of students and teachers is being formed to study the issue of student parking privileges. Fifteen staff members and 18 students have expressed an interest in serving on the committee. In how many different ways could a five person committee be formed if it must include at least one student and one teacher? 6. In the binary number system which is used in computer operations, there are only two digits allowed: 0 and 1. [ 225765 ] a) How many different binary numbers can be formed using at most four binary digits (ie: 0110) ? [ 16 ] b) If eight binary digits are used (ie: 11001101), how many different finery numbers can be formed? [ 256 ] Page 5 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec.4.1 An Introduction to Simulations and Experimental Probability A simulation is an experiment, model or activity that imitates real or hypothetical conditions. It is used to estimate quantities that are difficult to calculate and for verifying theoretical calculations. ie: biological experiments, drug testing, IKEA furniture testing, multiple choice guessing, births Example # 1: Simulating Birth Sequences Suppose a family plans to have four children. Use a simulation to estimate the likelihood that the family will have three girls in a row and then a boy. Solution # 1a) Toss a single coin four times to simulate a single trial. Let Heads represent the birth of a girl and let Tails represent the birth of a boy. Repeat this experiment for 50 trials to get a fair representation of the likelihood of having three girls followed by a boy. Solution # 1b) Toss four different coins to simulate the births of all four children at once. Let Heads represent the birth of a girl and let Tails represent the birth of a boy. The Penny will represent the 1st birth, the Nickel will represent the 2nd birth, the Dime will represent the 3rd birth and the Quarter will represent the 4th birth. Repeat this experiment for 50 trials to get a fair representation of the likelihood of having three girls followed by a boy. Example # 2: Batting Averages Suppose that Nicholas has a batting average of 0.320. This means that he will hit the ball 320 times out of 1000 at bats. This reduces to 32 out of 100 (or 8 out of 25). Use a simulation to estimate the likelihood that this player has no hits in one game (assuming 3 hits @ bat/game). Solution # 2 Fill a container with 25 slips of paper, 8 of which have the word HIT written on them. All the other papers will have the word MISS written on them, or you could leave them blank for saving time. Draw a sheet of paper out of the container and record “hit or miss”, return the paper to the container and draw again. You will draw 3 papers to represent Nicholas’s likelihood of hitting in a game. Repeat this experiment for 50 trials to get a fair representation of the likelihood that Nicholas will have no hits in a game. Page 6 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Using Graphing Calculators for Simulations The graphing calculator can be used for simulations. In Example # 1 for the births, you can generate a list of random numbers to represent the births of girls and boys. On the TI-83+ select the MATH button, use the right arrow 3 times so that PRB (for probability) is selected. Either scroll down or select #5 for RANDINT( , you now have to set a lower bound, upper bound and number of integers. In the case of example # 1, we will let our lower bound be 1 (which will represent the birth of a girl) and let our upper bound be 2 (which will represent the birth of a boy) and choose 4 because we wish to represent the births of four children in total. This function will generate a list of four numbers at random to simulate the birth of the four children. Repeat this experiment for 50 trials to get a fair representation of the birth sequences. ie: 2 2 1 2 (would represent a boy, boy, girl, boy for the four births) In the case of example # 2 we would need a random integer between 1 & 1000 with 3 numbers each time. The numbers 1 – 320 would represent HITS and the numbers 321 – 1000 would represent MISS. Repeat this experiment for 50 trials to find the likelihood that Nicholas will have no hits in a game. ie: 978 279 276 (would represent a miss, hit, hit during one game) Using Excel Spreadsheets for Simulations For Example # 1: Trial 1 2 3 4 5 6 7 8 9 10 Birth 1 Birth 2 Birth 3 Birth 4 2 1 1 2 1 1 2 2 1 1 1 2 1 1 2 1 2 1 1 1 2 2 2 1 1 2 2 1 1 1 1 1 1 1 2 1 2 1 1 1 Under the Birth 1 column you would enter “=int(2*rand( ) + 1). You then use the fill handle on this cell to drag this same formula into all other columns and rows necessary. Page 7 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec.4.2 Theoretical Probability Simple Event an event that consists of exactly one outcome. Experimental Probability only estimates the likelihood that an event occurs in a given experiment. Theoretical Probability is deduced from analysis of all possible outcomes. Sample Space is represented by “S” and is the collection of all possible outcomes. Event Space is represented by “A” and is the collection of all those outcomes that we want. We will look at the probability of experiments where the outcomes are easily determined. P A n A # of desired outcomes nS # of all possible outcomes Example #1: A die is rolled once. What is the probability of rolling a four? Solution # 1: Let S represent the collection of all possible outcomes. S = { 1 , 2 , 3 , 4 , 5 , 6 } Let A represent the successful outcomes. A = { 4 } Page 8 of 45 Prolling a four # of four total # Prolling a four 1 6 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 2: A bag contains 6 red smarties, 2 blue smarties, and 7 green smarties. If you take one, what is the probability that it is green? Solution # 2: Let S represent the collection of all smarties. S = { 15 smarties all together } Let A represent the green smarties. A = { 7 green } Pgreen smarties # of green total # Pgreen smarties 7 15 Example # 3: If a single die is rolled what is the probability of; a) rolling an odd number b) rolling a number less than 3 c) rolling either a three or a five Solution # 3: a) rolling an odd number Prolling an odd # # of odd numbers total # Prolling an odd # 3 6 1 Prolling an odd # 2 b) rolling a number less than 3 Prolling a # 3 # of numbers 3 total # Prolling a # 3 2 6 1 Prolling a # 3 3 Page 9 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master c) rolling either a three or a five Prolling a 3 or 5 # of 3 or 5 total # Prolling a 3 or 5 2 6 1 Prolling a 3 or 5 3 Example # 4: Find the probability that a number picked at random between 1 and 10 is a perfect square? Solution # 4: n { S } = 10 elements in total n { A } = 3 elements that are perfect squares between 1 & 10 n A n S 3 P# is perfect square 10 P# is perfect square Page 10 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 5: When drawing a single card from a standard deck of cards, what is the probability that; a) a queen is drawn b) a heart is drawn c) a black card is drawn Solution # 5: a) a queen is drawn n A n S 4 Pqueen 52 1 Pqueen 13 Pqueen b) a heart is drawn n A n S 13 P heart 52 1 P heart 4 P heart c) a black card is drawn n A n S 26 Pblack card 52 1 Pblack card 2 Pblack card Page 11 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master One immediate consequence of the definition of probabilities is the relationship between an event and it’s compliment, that is the set of outcomes in which the event does not occur. P A ' 1 P A In General: Example # 6: What is the probability of not selecting a prime number in a random selection of a number from 1 to 20? Solution # 6: Let A represent the prime numbers between 1 and 20. A = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } n(A)=8 & n ( S ) = 20 n A n S 8 P selecting a prime # 20 2 P selecting a prime # 5 P selecting a prime # Therefore, Pnot selecting a prime # 1 P A Pnot selecting a prime # 1 Pnot selecting a prime # Page 12 of 45 2 5 3 5 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 7: My music collection at home consists of eight rock CD’s, twelve blues CD’s and 4 classic CD’s. If a CD is chosen at random, find the probability that the one chosen is; a) rock b) not classical c) neither blues nor classical Solution # 7: a) rock n A n S 8 P rock 24 1 P rock 3 P rock b) not classical Pnot classical 1 Pclassical Pnot classical 1 4 24 Pnot classical 20 24 5 Pnot classical 6 c) neither blues nor classical Pneither blues nor classical Prock Pneither blues nor classical 8 24 1 Pneither blues nor classical 3 Page 13 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec.4.3a Finding Probability Using Sets A Set ie: is a collection or group of objects witch have a specific relationship. a set of dishes (all have the same pattern) the set of natural numbers (no fractions or decimals, 1 and up) An Element ie: is all the members of a set and are usually listed in curly brackets { } and each entry is separated by comma’s. D = { tea cup, saucer, bowl } this is called a finite set. n ( D ) = 3 N = { 1 , 2 , 3 , 4 , 5 , ………. } this is an infinite set n ( N ) = undefined Null Set is the set with no members in it. It is denoted by , or empty { } but not by{ 0 }. Disjoint Sets two sets that have no elements in common ie: C = { 4 , 6 , 7 , 9 , 10 } A = { 1 , 5 , 11 , 13 } C & A are disjoint sets. Subset ie: if all elements of one set are contained (also found in) a second set, then the first is a subset of the second. C={1,3,5,7} G={1,3,5,7,9} C is called a subset of G CG Compliment C ` is the compliment of C and contains all elements not in C but in the sample space S. Let our sample space be the odd #’s between 1 & 10 S 1 , 3 , 5 , 7 , 9 If C is defined as C = { 1 , 3 , 5 , 7 } the same as above, then C ` = { 9 } Page 14 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example #1: Describe the relationship between the sets described below using the terms equal, disjoint and subset. a) N = { the natural numbers } & W = { the whole numbers } b) R = { colours in a rainbow } & B = { grey, black, brown } c) D = { prime divisors of 34 } & E = { 2 , 17 } Solution # 1: a) N = { the natural numbers } & W = { the whole numbers } N W the natural numbers are contained within the whole numbers b) R = { colours in a rainbow } & B = { grey, black, brown } R & B are disjoint, no element is the same in the two sets. c) D = { prime divisors of 34 } & E = { 2 , 17 } D & E are equal because all the elements in D are also in E. All sets are subsets of a universal set for that particular situation ie: Page 15 of 45 the vowels V = { A , E , I , O , U } are a subset of the universal set called the alphabet V U where U = { A , B , C , D , E , F , G , …………, X , Y , Z } Prepared by: Scott McEwen MDM4U Unit 1 Master Let's say that our universe contains the numbers 1, 2, 3, and 4. Let A be the set containing the numbers 1 and 2. That is, A = {1, 2}. Let B be the set containing the numbers 2 and 3; that is, B = {2, 3}. Then we have the following relationships: The Union contains all the elements from both sets without repeats ie: A B means everything in A or B The Intersection contains all the elements that are common to both sets ie: A B means everything in A and B Recall the Compliment ie: A ` means everything in the universe not in A Try This? a) Find A B ` b) Find A B ` A B` A B` Page 16 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 2: Determine the elements in the union & intersection of M & N. M = { 1 , 3 , 6 , 9 , 12 } & N = { 2 , 4 , 6 , 8 , 10 , 12 } Solution # 2: Union of M & N is: M N { 1 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 } Intersection of M & N is : M N { 6 , 12 } Example # 3: Determine the elements in the following, given; S={K,E,Y,B,O,A,R,D} A={E,O,A} B={B,O,A,R,D} Find; a) A ` b) B ` c) A B d) A B e) A B` Solution # 3: a) A `= { K , Y , B , R , D } b) B ` = { K , E , Y } c) A B { O , A } d) A B { E , B , O , A , R , D } e) A B' { K , E ,Y , B , R , D } Page 17 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec.4.3b Finding Probability Using Venn Diagrams Try This? Example #1: Find the probability of rolling an even number or a number greater than 3 when rolling a die. These two events are dependent on each other. ie: there is overlap between the two events. Solution # 1: S={1,2,3,4,5,6} n(S) = 6 Let A be the event of rolling an even number. A={2,4,6} n(A) = 3 Let B be the event of rolling a number greater than 3 B={4,5,6} n(B) = 3 (A B) = { 4 , 6 } n(A B) = 2 P A B P A P B P A B 3 3 2 6 6 6 2 3 *** you must take away the intersection *** Now let’s look at a situation where the two events are not dependent on each other. They are independent or disjointed events, sometimes also referred to as mutually exclusive. Page 18 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 2: If two dice are rolled, one red and one green (so that you can tell the difference between them), find the probability that a sum of; a) 3 or 12 will occur Solution # 2a Let A be the event of rolling a sum of 3 Let B be the event of rolling a sum of 12 ** These two events are mutually exclusive, independent ** P A P B 2 36 Dice # 1 Sums of two dice 1 2 3 4 5 6 b) P A B P A P B 1 36 1 2 Dice # 2 3 4 5 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 2 1 36 36 3 36 1 12 6 7 8 9 10 11 12 6 or a pair will occur Solution # 2b Let A be the event of rolling a sum of 6 Let B be the event of rolling a pair ** These two events are not mutually exclusive, they are dependent because there is 6 in both sets! ** P A 5 36 P B 6 36 P A B 1 36 ** account for 6 in both ** P A B P A P B P A B 5 6 1 36 36 36 10 36 5 18 Page 19 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Mutually exclusive events occur when 2 events can not happen at the same time. B A ** there is nothing to take away when the two sets are disjoint ** P A B P A PB Example # 3: Jody attends a fundraiser at which T-shirts are given away as door prizes. Door prize winners are randomly given a shirt from a stock of 2 black shirts, 4 blue shirts and 9 white shirts. Assuming that Jody wins the first door prize what is the probability that she will get a black or white shirt? Solution # 3: Let A be the event of drawing a black shirt Let B be the event of drawing a white shirt ** these are mutually exclusive, independent events ** P A 2 15 P B 9 15 P A B P A P B 2 9 15 15 11 15 Page 20 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 4: A card is randomly selected from a standard deck of cards. What is the probability that either a heart or a face card (JQK) is selected. Solution # 4: Let A be the event of selecting a heart Let B be the event of selecting a face card ** these are not mutually exclusive, they are dependent events ** P A P B 13 52 12 52 P A B 3 52 ** J of heart, Q or heart, K of hearts ** P A B P A P B P A B 13 12 3 52 52 52 22 52 11 26 Example # 5: As a result of a recent survey it was estimated that 85% of the grade twelve population at a local high school enjoys an alcoholic beverage once a week, 35% smoke at least one cigarette a day and 25% indulge in both habits. What is the probability that an individual chosen at random from the targeted population either smokes or drinks. Solution # 5: Let A be the event that the individual drinks Let B be the event that the individual smokes ** these are not mutually exclusive, they are dependent events ** P(A) = 0.85 P(B) = 0.35 P(A B) = 0.25 P A B P A P B P A B 0.85 0.35 0.25 0.95 Therefore the probability that an individual either smokes or drinks when chosen at random from the targeted population is 95 %. Page 21 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.4 Conditional Probability When calculating probabilities, additional information may become available that will affect the calculation of the outcome. ie: If you are interested in the probability of your passing the next test, the probability will depend on whether you studied, if you worked late the night before, if you had a good breakfast, have you been doing your homework regularly, etc… When additional information is known that will affect the probability of an outcome, we calculate what is called the conditional probability represented by P(A | B), read as “the probability of event A, given that the event B has occurred”, or, “the probability of A given B”. P A | B P A B P B PB | A P B A P A **N.B.** The P A B is identical to the PB A Example #1: What is the probability of rolling a sum greater than 7 with two dice if it is known that the first die rolled is a 3. Solution # 1: Let A be the event of a sum greater than 7. Let B be the event of the first roll being a 3. P A B P B 2 6 36 36 2 6 1 3 P A | B 1 The probability of rolling a sum greater than 7, given that the first roll was a 3 is 3 , 0.33333 or 33.3% We can rearrange the conditional probability formula to find the intersection of two events if the conditional probability is known. P A | B P A B P B P A B PB P A | B Page 22 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 2: When drawing two cards from a standard deck of 52 cards (without replacement) what is the probability that they are both aces? Solution # 2 Let A be the event that the first card drawn is an ace. Let B be the event that the second card drawn is an ace. P B A P A B P A P B | A 4 3 52 51 1 1 13 17 1 221 1 Therefore there is a 221 or approx. 0.45% chance that both aces will be drawn. Example # 3: A professional hockey team has 8 wingers. Three of these wingers are 30-goal scorers or “snipers”. Every fall the team plays a match with the club’s farm team. The coaches agree to select two wingers at random from the pro-team to play for the farm team. What is the probability that two snipers will play for the farm team. Solution # 3: Let A be the event the first winger chosen is a sniper. Let B represent the second winger chosen is a sniper. P B A P A B P A P B | A 3 2 8 7 3 1 4 7 3 28 3 Therefore there is a 28 or approx. 10.71% chance of selecting two wingers that are both snipers. Page 23 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 4: 2 1 The probability that Ann will go to Queens is 5 . The probability that she will go to another university is 2 . 3 If Ann goes to Queens, the probability that her boyfriend, James, will follow her and go to Queens is 4 . What is the probability that both Ann and James will attend Queens? Solution # 4: Let A be the event that Ann goes to Queens. Let B be the event that James goes to Queens. P B A P A B P A P B | A 2 3 5 4 1 3 5 2 3 10 3 Therefore there is a 10 or approx. 30% chance that both will attend Queens. Example # 5: A brown bag contains 2 white and 2 red balls. A ball is drawn at random. If the ball chosen is white, then it is returned to the bag along with an extra white ball. If the ball chosen on the first draw is red, then it is returned to the bag with 2 extra red balls. Then a second ball is drawn from the bag. Find the probability that the second ball drawn is red? Solution # 5: Let R represent choosing a red ball. Let W represent choosing a white ball. **drawing a tree diagram will help sort out the different probabilities** Pred 2nd pick Pred & red or Pwhite & red 1 4 1 2 2 6 2 5 1 1 3 5 8 15 8 Therefore the probability of getting a red ball on the second drawn is 15 or approx. 53.3% Page 24 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec.4.5 Tree Diagrams & Dependent vs. Independent Events Independent Events the probability of one event happening is not affected by the outcome of the other events. Example #1: Consider an experiment in which you roll a 6 sided dice and then toss a coin. What is the probability of tossing tails and rolling and even number. Solution # 1: Let E be the event of rolling an even number. Let T be the event of tossing tails. We can create an outcomes table to show all of the possibilities for these particular events. Roll Toss 1 H 1 T 2 H 2 T 3 H 3 T 4 H 4 T 5 H 5 T 6 H 6 T Page 25 of 45 neven & tail n(total ) 3 12 1 4 Peven & tail Peven & tail Peven Ptail 3 1 6 2 1 4 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 2: (Review of Conditional Probability) A card is drawn from a deck of cards. What is the probability that it is a jack given that it is a face card? Solution # 2 Let A be the even of drawing a Jack. Let B be the event of drawing a face card. P A B P B 4 52 12 52 4 52 52 12 1 3 P A | B Therefore there is a 33.3% chance of drawing a jack knowing that the card is a face card. Recall, when 2 events have no effect on one another they are independent events. P A B P A P B Example # 3: Justin estimates that his probability of passing English is 0.9 and his probability of passing Math is 0.8. Find the probability that Justin will; a) pass both Math and English b) pass Math but fail English c) not pass Math or English Solution # 3: Let A be the event of passing Math. a) Let B be the event of passing English. P A B P A P B 0 .8 0 .9 0.72 Therefore there is a 72% chance of passing both Math & English. Page 26 of 45 Prepared by: Scott McEwen MDM4U b) Unit 1 Master P A B ' P A P B ' 0 .8 0 .1 0.08 Therefore there is an 8% chance of passing Math but failing English. c) P A ' B ' P A ' PB ' 0 .2 0 .1 0.02 Therefore there is a 2% chance of not passing Math or English. Example # 4: 4 George estimates that the probability of getting the next guess right if the previous one was right on a test is 5 . 2 But the probability of getting it right if the previous one was wrong is only 5 . If the probability of getting her 3 first guess right is 4 , what is the probability of getting the second guess right? Solution # 4: Let A be the event that the first is correct. Let B be the event that the second is correct. **Draw a tree diagram to organize your probabilities** P2nd is correct P A B P A ' B 12 2 20 20 14 20 7 10 Therefore there is a 70% chance of getting the second guess correct. Page 27 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.6a Factorials & Permutations Example #1: In how many ways can you arrange 7 smarties that are all different colors. Solution # 1: 7 6 5 4 3 2 1 = 7 ! = 5040 Factorials n! nn 1n 2n 3........ 3 2 1 Example # 2: Evaluate each of the following factorial expressions. a) 9! d) 8! 6! b) 0! e) 75 ! 71 ! c) 1! f) 17 ! 15 ! 2 ! Solution # 2 a) 9 ! = 362 880 d) 8 7 6 5 4 3 2 1 6 5 4 3 2 1 56 Page 28 of 45 b) 0 ! = 1 e) 75 74 73 72 71 70 ......... 71 70 ......... 29,170,800 c) 1 ! = 1 f) 17 16 15! 15! 2! 132 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 3: a) Albert, Jo, Kim and Jane are to make speeches in English class. How many different arrangements are there for who will go 1st, 2nd, 3rd and 4th ? b) If the English class has 10 students to present but only has time for 4 students how many possible arrangements are there? Solution # 3: a) 4 3 2 1 = 4 ! = 24 b) 10 9 8 7 = 10P4 = 10! 10! 5040 10 4! 6! A permutation is the number of different arrangements that can be made from a selection of objects in a definite order. In general, if we have “n” distinct objects and we want to pick “r” of them to arrange, the number of possible n! different arrangements is n r ! , denoted by P(n,r) or nPr. Example # 4: An investment club of 5 members wants to select a president and vice-president. In how many ways can this be done? Solution # 4: Choosing 2 people from 5 where order matters. Therefore, 5P2 or P(5,2) 5! 5! 5 4 20 5 2 ! 3 ! Page 29 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 5: a) Find the number of permutations of the letters of the word “DIPLOMA” for which the letter “L” must be in the middle? b) How many ways are there of arranging the letters of the word “DIPLOMA” so that the “O” and “I” are together? Solution # 5: a) 6 5 4 L 3 2 1 = 6 ! = 720 b) treat “OI” as a boxed item, there are 6 ! ways of arranging that box and then 2 ! ways of arranging the “OI” inside the box. Therefore, 6 ! 2 ! = 1440 Example # 6: At a used car lot, six different model cars are lined up at the front of the lot. In how many ways can this be done if: a) the only car with a sunroof must be at the right end of the line? b) the three black cars must be together? Solution # 6: a) 5 4 3 2 1 Sunroof = 5 ! = 120 b) “3 black cars” 3 2 1 = 4 ! 3 ! = 144 Page 30 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Arrangement of objects when some are alike Example # 7: Consider taking a photograph of three friends, Joel, Josh & Lisa. a) In how many ways can the friends be lined up? b) Now consider that Joel & Josh are identical twins. How many different ways would the photo look to people who do not know the twins? Solution # 7: a) 3 ! = 6 b) Joel Josh Lisa JJL Josh Joel Lisa JJL Joel Lisa Josh JLJ Josh Lisa Joel JLJ Lisa Joel Josh LJJ Lisa Josh Joel LJJ There are only 3 distinct ways to view the photo. 3! 3 2! In general, “n” objects, of which some are alike, can be arranged in: n! a !b!c ! Example # 8: In how many ways can the letters of the word “MINIMUM” be arranged? Solution # 8: 7! 420 3 ! 2 ! a) Page 31 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 9: Anya is starting out on her evening run. Her route always takes her 8 blocks east and five blocks north to her grandmothers house. She likes to vary her path every night. How many different possibilities are there? Solution # 9: 13 ! 1287 a) 8 ! 5 ! Example # 10: Determine the number of arrangements possible using all the letters of the word “PINEAPPLE” if; a) there are no restrictions b) the “N” must go first c) the “A” must be first and the “N” must be last Solution # 10: a) 9! 30240 3!2! b) Place the “N” and arrange the other 8 letters. Therefore, 1 8 ! 3360 3!2! c) Place the “A” & “N” and arrange the other 7 letters. Therefore, 1 7 ! 420 3!2! Page 32 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.6b Probabilities and Permutations Since, P event nevent set noutcome set nwith restrictio ns nwithout restrictio ns Example #1: When the kindergarden class of 18 students is lining up, what is the probability that Beth is first? Solution #1: n(Beth first) = 17 ! PBeth 1st n(no restrictions) = 18 ! 17 ! 1 18 ! 18 Example #2: In making a 7 digit number by arranging the digits 1 , 2 , 2 , 3 , 5 , 5 , 6 what is the probability of getting a number less than 3 000 000? Solution #2: 3 6 5 4 3 2 1 nrestrictions 3 6 ! 2!2! nwithout restrictio ns 7! 2!2! 3 6 ! 2!2! 3 P 3,000,000 7! 7 2!2! Page 33 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example #3: When the 5 starting skaters on the hockey team line up at the blue line, what is the probability that the two defence men are not beside each other? Solution #3: n(2 defence men are together) = 4 3 2 1 x 2 ! Ptogether nwithout restrictions 5 ! 2 ! 4 ! 2 5! 5 Pnot together 1 Ptogether 1 2 5 3 5 Example #4: In how many ways can a President, Vice-President and a Secretary be arranged around a circular table? Solution #4: For every 3 different arrangements in a line, there is only 1 corresponding arrangement in a circle. Therefore, you have to fix one person around a circular table to get a new arrangement. 3! 2 3 In General, # of ways Page 34 of 45 n! n 1! n Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.7a Combinations A selection from a group of items without regard to order is called a combination. Example #1: List all the two letter permutations that can be made using the letters “A , B , C” without repetition. Solution # 1: AB AC BC BA CA CB Therefore, the # of different permutations is 3 ! = 6 Example # 2: Now, a construction crew of two people is to be chosen from Al, Ben and Charlie. List all possible groupings. Solution # 2 AB AC BC Therefore, the # of different groups is; 3 3! 2! Since arranging the two people once selected does not matter. The combinations of “n” distinct objects taken “r” at a time is a selection of “r” of these objects without regard to order. n n! C n, r n C r r n r ! r ! Notation: Page 35 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 3: Evaluate the following combinations; a) 8 3 a) 8 56 3 b) 8 5 b) 8 56 5 Solution # 3: Why are they the same (using the definition of combinations and factorials)? Connection to Pascal’s Triangle? Example # 4: From 20 members of the parent council; a) how many ways can you choose a president, vice-president and a secretary? b) in how many ways can you choose three people to bring cookies? Solution # 4: 20 P3 a) 20 ! 20 19 18 6840 20 3! 20 b) C3 20 ! 20 19 18 1140 20 3! 3 ! 3 2 Example # 5: To play lotto 649, you must select 6 different numbers from 1 to 49 (the order of the numbers does not matter). How many ways can this be done? Solution # 5: 49 C6 49 ! 13,983,816 49 6! 6 ! Page 36 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 6: Joan runs a small landscaping business. She has on hand 12 kinds of rose bushes, 16 kinds of small shrubs, 11 kinds of evergreen seedlings and 18 kinds of flowering lilies. In how many ways can Joan fill an order if a customer wants; a) 15 different varieties consisting of 4 roses, 3 shrubs, 2 evergreens and 6 lilies? b) Either 4 different roses or 6 different lilies? Solution # 6a) Roses = 12 C4 and Therefore, 12 Shrubs = 16 C3 and Evergreens = 11 C2 and Lilies = 18 C6 C 4 16 C3 11 C 2 18 C 6 2.83 1011 Solution # 6b) 12 Roses = 12 C4 or Lilies = 18 C 6 Therefore, C 4 18 C 6 495 18564 19059 Example # 7: A delegation of three people is to be chosen from a group of community volunteers consisting of four lawyers, a minister, and three retail merchants. In how many ways can this group be formed if at least one retailer must be a member of the delegation? Solution # 7a Direct Reasoning (consider 3 cases) i) 1 retailer, or ii) 2 retailers, or iii) 3 retailers 3 5 3 10 30 1 2 3 5 3 5 15 2 1 3 5 1 1 1 3 0 Therefore, # of ways is 30 + 15 + 1 = 46 Solution # 7b Indirect Reasoning (find the total # of 3 person groups without restrictions, then subtract the # with no retailers) 8 5 46 3 3 Page 37 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 8: There are 8 boys and 12 girls in a drama class. How many ways can a committee of 5 be selected in each case; a) There must be exactly 2 boys and 3 girls? b) There must be at least 2 boys? Solution # 8: 8 12 28 220 6160 a) 2 3 b) Direct Reasoning; 2 boys + 3 boys + 4 boys +5 boys Indirect Reasoning All – 1 boy – no boys 20 no restrictions 15504 5 8 12 1 boy 3960 1 4 12 no boy 792 5 Therefore, 15504 – 3960 – 792 = 10752 Page 38 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.7b Probability & Odds Try This? Example #1: When drawing two cards from a regular deck of 52 cards without replacement, what is the probability that they are both aces? Solution # 1a: 4 3 Logically the chance for the first ace is 52 and the chance for the second ace is 51 . As seen in our previous examples from the other day, this is approx. 0.0045 Solution # 1b: Using combinations Ptwo aces # of combinations with restrictio ns # of combinations without restrictio ns 4 2 0.0045 52 2 Example #2: From a class of 12 boys and 11 girls, what is the probability that a group of 5 people chosen at random contains: a) exactly 3 girls Solution # 2a a) 1112 3 2 23 5 0.3236 b) at least one boy Solution # 2b Pat least1 b) 1 Pno boys 1112 5 0 1 23 5 1 0.01373 0.98627 Page 39 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master ODDS 1 We learned that the probability of rolling a 6 in a single roll of a die is 6 . The odds of rolling a six is 1:5 . The odds against rolling a six are 5:1. The odds for (in favour) of an event A can be expressed as the ratio P(A):P(A’), remember that P(A’) = 1 – P(A). P A ' P A odds in favour of A odds against A P A ' P A Example #3: A messy drawer contains three red socks, five white socks and four black socks. What are the odds in favour of randomly drawing a red sock? Solution # 3: Let A be the event of drawing a red sock. P A P A' 1 3 1 12 4 1 3 4 4 P A P A' 1 4 3 4 1 4 1 4 3 3 odds in favour Therefore the odds in favour of drawing a red sock are 1:3 Example #4: If the chance of a snowstorm in Windsor, in January is estimated at 0.4, what are the odds against Windsor having a snowstorm next January? Solution # 4: Let A be the event of a snowstorm in January. P(A) = 0.4 P(A’) = 0.6 odds against 0.6 3 0.4 2 Odds against a snow storm are 3:2. Since this is greater than a 1:1 ratio, snow is less likely to occur. Page 40 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example # 5: A university professor, in an effort to promote good attendance habits, states that the odds of passing her course are 8:1 when a student misses fewer than five classes. What is the probability that a student with good attendance will pass? Solution # 5: Let A be the event that a student with good attendance passes. odds in favour of A P A P A' 8 P A 1 1 P A 8 8 P A P A P A 8 9 Therefore the odds in favour of a student with good attendance passing is 8:9 Example # 6: In choosing two cards from a deck of 52, what are the odds in favour of both cards being black? Solution #6: Pboth black 26 2 52 2 25 102 25 odds in favour 102 77 102 25 77 Therefore the odds in favour of both cards chosen being black is 25:77 Page 41 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Example #7: If drawing 5 cards from a deck of 52, what are the odds in favour of a full house? Solution # 7: P full house 4 4 13 12 3 2 52 5 3744 2,598,960 6 4165 6 odds in favour 4165 4159 4165 6 4159 Therefore there is a 6:4159 odds in favour of getting a full house. Page 42 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master Sec. 4.7c Problem Solving with Combinations Example #1: Twenty people are to travel in a bus from the airport to the hotel at a resort. The bus is designed for use in tropical climate, it can carry 12 passengers outside and eight inside. If four of the passengers refuse to travel outside and five will not travel inside, in how many ways can the passengers be seated if the demands of the passengers are all met? Solution # 1: Seat all the picky people first. Inside – 4 must be, 4 left Outside – 5 must be, 7 left Therefore, 9 already seated, 11 left to sit! 11 4 330 7 4 outside inside Example # 2: An artist has an apple, an orange and a pear in his refrigerator. In how many ways can the artist choose one or more piece of fruit for a still life painting? Solution # 2 Artist has 2 choices for each piece of fruit. Include or Not! # of choices = 2 x 2 x 2 = 8 But not choosing each will leave nothing to paint. We need to have at least one piece of fruit. Therefore, 23 – 1 = 7 Page 43 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master The total number of combinations containing at least one item chosen from a group of “n” distinct items is 2n – 1. If we want to include the null set we need to use just 2n. Example # 3: In how many ways can you form a committee if you have a delegate of six members? Solution # 3: 6 6 6 6 6 6 6 2 6 0 1 2 3 4 5 6 However, you can’t have a committee with 0 people or one person therefore the first two combinations don’t count. 26 – 1 – 6 = 64 – 1 – 6 = 57 Therefore, there is 57 ways to choose at least one person from a 6 member delegate. All possible combinations with some identical items Example # 4: Katie is responsible for stocking the coffee room at her office. She can purchase up to 3 cases of cookies, four cases of soft drinks and two cases of coffee packets without having to send the order through accounting. How many different direct purchases can Katie make? Solution # 4: Cookies 0 1 2 3 3 4 Soft drinks 0 1 2 Coffee 0 1 2 = (Cookies x Soft Drinks x Coffee) – 1 = (4 x 5 x 3) – 1 = 59 Page 44 of 45 Prepared by: Scott McEwen MDM4U Unit 1 Master If at least one item is chosen, the total number of selections that can be made from “p” items of one kind, “q” items of another kind, etc……. is (p+1)(q+1)……- 1 Example # 5: Patti wants to throw coins in the fountain at the mall and make a wish. a) She has a penny, a nickel, a dime and a quarter in her pocket. How many different sums of money can she throw into the fountain? b) How many different sums of money could she throw if she had three more pennies? Solution # 5: a) P (2 b) x P (5 Page 45 of 45 N 2 D x N x 2 2 Q x D x 2 2) -1= 15 -1= 39 Q x 2) Prepared by: Scott McEwen