#### Transcript 12 Probability

Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION Slide 12-1 Chapter 12 Probability Copyright © 2005 Pearson Education, Inc. 12.1 The Nature of Probability Copyright © 2005 Pearson Education, Inc. Definitions An experiment is a controlled operation that yields a set of results. The possible results of an experiment are called its outcomes. An event is a subcollection of the outcomes of an experiment. Copyright © 2005 Pearson Education, Inc. Slide 12-4 Definitions continued Empirical probability is the relative frequency of occurrence of an event and is determined by actual observations of an experiment. Theoretical probability is determined through a study of the possible outcome that can occur for the given experiment. Copyright © 2005 Pearson Education, Inc. Slide 12-5 Empirical Probability P (E ) number of times event E has occurred total number of times the experiment has been performed Example: In 100 tosses of a fair die, 19 landed showing a 3. Find the empirical probability of the die landing showing a 3. Let E be the event of the die landing showing a 3. P (E ) Copyright © 2005 Pearson Education, Inc. 19 0.19 100 Slide 12-6 The Law of Large Numbers The law of large numbers states that probability statements apply in practice to a large number of trials, not to a single trial. It is the relative frequency over the long run that is accurately predictable, not individual events or precise totals. Copyright © 2005 Pearson Education, Inc. Slide 12-7 12.2 Theoretical Probability Copyright © 2005 Pearson Education, Inc. Equally likely outcomes If each outcome of an experiment has the same chance of occurring as any other outcome, they are said to be equally likely outcomes. P (E ) number of outcomes favorable to E total number of outcomes Copyright © 2005 Pearson Education, Inc. Slide 12-9 Example A die is rolled. Find the probability of rolling a) a 3. b) an odd number c) a number less than 4 d) a 8. e) a number less than 9. Copyright © 2005 Pearson Education, Inc. Slide 12-10 Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. a) P (2) number of outcomes that will result in a 2 1 total number of possible outcomes b) There are three ways an odd number can occur 1, 3 or 5. 3 1 P (odd) 6 6 2 c) Three numbers are less than 4. 3 1 P (number less than 4) 6 2 Copyright © 2005 Pearson Education, Inc. Slide 12-11 Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. continued d) There are no outcomes that will result in an 8. 0 P (number greater than 8) 6 0 e) All outcomes are less than 10. The event must occur and the probability is 1. Copyright © 2005 Pearson Education, Inc. Slide 12-12 Important Facts The probability of an event that cannot occur is 0. The probability of an event that must occur is 1. Every probability is a number between 0 and 1 inclusive; that is, 0 P(E) 1. The sum of the probabilities of all possible outcomes of an experiment is 1. Copyright © 2005 Pearson Education, Inc. Slide 12-13 Example A standard deck of cards is well shuffled. Find the probability that the card is selected. a) a 10. b) not a 10. c) a heart. d) a ace, one or 2. e) diamond and spade. f) a card greater than 4 and less than 7. Copyright © 2005 Pearson Education, Inc. Slide 12-14 Example continued a) a 10 There are four 10’s in a deck of 52 cards. 4 1 P (10) 52 13 Copyright © 2005 Pearson Education, Inc. b) not a 10 P (not a 10) 1 P (10) 1 1 13 12 13 Slide 12-15 Example continued c) a heart d) an ace, 1 or 2 There are 13 hearts in the deck. There are 4 aces, 4 ones and 4 twos, or a total of 12 cards. 13 1 P (heart) 52 4 Copyright © 2005 Pearson Education, Inc. 12 3 P (A, 1 or 2) 52 13 Slide 12-16 Example continued d) diamond and spade e) a card greater than 4 and less than 7 The word and means both events must occur. This is not possible. The cards greater than 4 and less than 7 are 5’s, and 6’s. 0 P (diamond & spade) 0 52 Copyright © 2005 Pearson Education, Inc. 8 2 P (E ) 52 13 Slide 12-17 12.3 Odds Copyright © 2005 Pearson Education, Inc. Odds Against Odds against event = P (event fails to occur) P (failure) P (event occurs) P (success ) Example: Find the odds against rolling a 5 on one roll of a die. 1 P (5) 6 P(odds) 5 P (fails to roll a 5) 6 5 6 1 6 5 6 5 6 1 1 Copyright © 2005 Pearson Education, Inc. The odds against rolling a 5 are 5:1. Slide 12-19 Odds in Favor P (event occurs) Odds in favor of event = P (event fails to occur) P (success ) P (failure) Copyright © 2005 Pearson Education, Inc. Slide 12-20 Example Find the odds in favor of landing on blue in one spin of the spinner. 3 P (blue)= 8 5 P (not blue)= 8 P(odds in favor) 3 8 5 8 3 8 3 8 5 5 The odds in favor of spinning blue are 3:5. Copyright © 2005 Pearson Education, Inc. Slide 12-21 Probability from Odds Example: The odds against spinning a blue on a certain spinner are 4:3. Find the probability that a) a blue is spun. b) a blue is not spun. Copyright © 2005 Pearson Education, Inc. Slide 12-22 Solution Since the odds are 4:3 the denominators must be 4 + 3 = 7. The probabilities ratios are: Copyright © 2005 Pearson Education, Inc. 4 P (blue)= 7 3 P (not blue)= 7 Slide 12-23 12.4 Expected Value (Expectation) Copyright © 2005 Pearson Education, Inc. Expected Value E = x1P(x1) + x2P(x2) + x3P(x3) +…+ xnP(xn) or E = P(x1)x1 + P(x2)x2 + P(x3)x3 +…+ P(xn)xn (order doesn’t matter) The symbol P1 represents the probability that the first event will occur, and A1 represents the net amount won or lost if the first event occurs. Copyright © 2005 Pearson Education, Inc. Slide 12-25 Example Teresa is taking a multiple-choice test in which there are four possible answers for each question. The instructor indicated that she will be awarded 3 points for each correct answer and she will lose 1 point for each incorrect answer and no points will be awarded or subtracted for answers left blank. If Teresa does not know the correct answer to a question, is it to her advantage or disadvantage to guess? If she can eliminate one of the possible choices, is it to her advantage or disadvantage to guess at the answer? Copyright © 2005 Pearson Education, Inc. Slide 12-26 Solution Expected value if Teresa guesses. 1 P (guesses correctly) 4 3 P (guesses incorrectly) 4 1 3 Teresa's expectation = (3) ( 1) 4 4 3 3 0 4 4 Copyright © 2005 Pearson Education, Inc. Slide 12-27 Solution continued—eliminate a choice 1 P (guesses correctly) 3 2 P (guesses incorrectly) 3 1 2 Teresa's expectation = (3) ( 1) 3 3 2 1 1 3 3 Copyright © 2005 Pearson Education, Inc. Slide 12-28 Example: Winning a Prize When Calvin Winters attends a tree farm event, he is given a free ticket for the $75 door prize. A total of 150 tickets will be given out. Determine his expectation of winning the door prize. 1 149 Expectation = (75) (0) 150 150 1 2 Copyright © 2005 Pearson Education, Inc. Slide 12-29 Example When Calvin Winters attends a tree farm event, he is given the opportunity to purchase a ticket for the $75 door prize. The cost of the ticket is $3, and 150 tickets will be sold. Determine Calvin’s expectation if he purchases one ticket. Copyright © 2005 Pearson Education, Inc. Slide 12-30 Solution 1 149 E ( x) 72 3 150 150 72 447 150 150 375 150 2.5 Calvin’s expectation is $2.50 when he purchases one ticket. Copyright © 2005 Pearson Education, Inc. Slide 12-31 Fair Price Fair price = expected value + cost to play Copyright © 2005 Pearson Education, Inc. Slide 12-32 Example Suppose you are playing a game in which you spin the pointer shown in the figure, and you are awarded the amount shown under the pointer. If is costs $10 to play the game, determine a) the expectation of the person who plays the $10 $2 game. $15 $20 b) the fair price to play the game. $10 $2 Copyright © 2005 Pearson Education, Inc. Slide 12-33 Solution Amt. Shown on Wheel $2 $10 $15 $20 Probability 3/8 3/8 1/8 1/8 Amt. Won/Lost $8 $0 $5 $10 3 3 1 1 Expectation = ( $8) ($0) ($5) ($10) 8 8 8 8 24 5 10 0 8 8 8 9 1.125 1.13 8 Copyright © 2005 Pearson Education, Inc. Slide 12-34 Solution Fair price = expectation + cost to play = $1.13 + $10 = $8.87 Thus, the fair price is about $8.87. Copyright © 2005 Pearson Education, Inc. Slide 12-35 12.5 Tree Diagrams Copyright © 2005 Pearson Education, Inc. Counting Principle If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct way, then the two experiments in that specific order can be performed in M • N ways. (Note: Order is important.) Copyright © 2005 Pearson Education, Inc. Slide 12-37 Definitions Sample space: A list of all possible outcomes of an experiment. Sample point (simple event): Each individual outcome in the sample space. Copyright © 2005 Pearson Education, Inc. Slide 12-38 Example Two balls are to be selected without replacement from a bag that contains one purple, one blue, and one green ball. a) Use the counting principle to determine the number of points in the sample space. b) Construct a tree diagram and list the sample space. c) Find the probability that one blue ball is selected. d) Find the probability that a purple ball followed by a green ball is selected. Copyright © 2005 Pearson Education, Inc. Slide 12-39 Solutions a) 3 • 2 = 6 ways b) P B G B G P G PB PG 4 2 c) P (blue)= 6 3 d) P(purple then green = 1/6 BP BG P GP B GB Note: The probability of each simple event is 1 1 1 . 3 2 6 Copyright © 2005 Pearson Education, Inc. Slide 12-40 12.6 Or and And Problems Copyright © 2005 Pearson Education, Inc. Or Problems P(A or B) = P(A) + P(B) P(A and B) Example: Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains an even number or a number greater than 5. Copyright © 2005 Pearson Education, Inc. Slide 12-42 Solution P(A or B) = P(A) + P(B) P(A and B) even or even and P P (even) P (greater 5) greater than 5 greater than 5 5 5 3 10 10 10 7 10 Thus, the probability of selecting an even number or a number greater than 5 is 7/10. Copyright © 2005 Pearson Education, Inc. Slide 12-43 Example Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains a number less than 3 or a number greater than 7. Copyright © 2005 Pearson Education, Inc. Slide 12-44 Solution 2 P (less than 3) 10 3 P (greater than 7) 10 There are no numbers that are both less than 3 and greater than 7. Therefore, 2 10 3 5 1 0 10 10 2 Copyright © 2005 Pearson Education, Inc. Slide 12-45 Mutually Exclusive Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously. Copyright © 2005 Pearson Education, Inc. Slide 12-46 Example One card is selected from a standard deck of playing cards. Determine the probability of the following events. a) selecting a 3 or a jack b) selecting a jack or a heart c) selecting a picture card or a red card d) selecting a red card or a black card Copyright © 2005 Pearson Education, Inc. Slide 12-47 Solutions a) 3 or a jack b) jack or a heart 4 4 52 52 8 2 52 13 P (3) P ( jack) jack and 4 13 1 P (jack) P (heart) P heart 52 52 52 16 4 52 13 Copyright © 2005 Pearson Education, Inc. Slide 12-48 Solutions continued c) picture card or red card picture & 12 26 6 P (picture) P (red) P red card 52 52 52 32 8 52 13 d) red card or black card 26 26 P (red) P (black) 52 52 52 1 52 Copyright © 2005 Pearson Education, Inc. Slide 12-49 And Problems P(A and B) = P(A) • P(B) Example: Two cards are to be selected with replacement from a deck of cards. Find the probability that two red cards will be selected. P ( A) P (B ) P (red ) P (red ) 26 26 52 52 1 1 1 2 2 4 Copyright © 2005 Pearson Education, Inc. Slide 12-50 Example Two cards are to be selected without replacement from a deck of cards. Find the probability that two red cards will be selected. P (red and red ) P(red ) P(red ) 26 25 650 25 52 51 2652 102 Copyright © 2005 Pearson Education, Inc. Slide 12-51 Independent Events Event A and Event B are independent events if the occurrence of either event in no way affects the probability of the occurrence of the other event. Experiments done with replacement will result in independent events, and those done without replacement will result in dependent events. Copyright © 2005 Pearson Education, Inc. Slide 12-52 Example A package of 30 tulip bulbs contains 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. Three bulbs are randomly selected and planted. Find the probability of each of the following. All three bulbs will produce pink flowers. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower. None of the bulbs will produce a yellow flower. At least one will produce yellow flowers. Copyright © 2005 Pearson Education, Inc. Slide 12-53 Solution 30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. All three bulbs will produce pink flowers. P 3 pink P (pink 1) P (pink 2) P (pink 3) 6 5 4 = 30 29 28 1 203 Copyright © 2005 Pearson Education, Inc. Slide 12-54 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower. P red,yellow,red P (red) P (yellow) P (red) 14 10 13 30 29 28 13 174 = Copyright © 2005 Pearson Education, Inc. Slide 12-55 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. None of the bulbs will produce a yellow flower. first not second not third not P none yellow P P P yellow yellow yellow 20 19 18 = 30 29 28 57 203 Copyright © 2005 Pearson Education, Inc. Slide 12-56 Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. At least one will produce yellow flowers. P(at least one yellow) = 1 P(no yellow) 57 1 203 146 203 Copyright © 2005 Pearson Education, Inc. Slide 12-57 12.7 Conditional Probability Copyright © 2005 Pearson Education, Inc. Conditional Probability In general, the probability of event E2 occurring, given that an event E1 has happened is called conditional probability and is written P(E2|E1). Copyright © 2005 Pearson Education, Inc. Slide 12-59 Example Given a family of two children, and assuming that boys and girls are equally likely, find the probability that the family has a) two girls. b) two girls if you know that at least one of the children is a girl. c) two girls given that the older child is a girl. Copyright © 2005 Pearson Education, Inc. Slide 12-60 Solutions a) two girls There are four possible outcomes BB, BG, GB, and GG. 1 P (2 girls) = 4 b) two girls if you know that at least one of the children is a girl 1 P (both girls|at least one is a girl) = 3 Copyright © 2005 Pearson Education, Inc. Slide 12-61 Solutions continued two girls given that the older child is a girl S = {BB, BG, GB, GG} Older is a Girl BG, GG Two girls given older is a girl GG 1 P (both girls|older child is a girl) = 2 Copyright © 2005 Pearson Education, Inc. Slide 12-62 Conditional Probability For any two events, E1, and E2, P( E2 | E1 ) n( E2 AND E1 ) n( E1 AND E2 ) n( E1 ) n( E1 ) P( E1 AND E2 ) P( Both Events) P( E1 ) P(Given Event) Exam ple: P (2 girlsE2 | olderis a girlE1 ) n(2 girlsE2 AND olderis a girlE1 ) n(olderis a girlE1 ) 1 2 Copyright © 2005 Pearson Education, Inc. S = {BB, BG, GB, GG} Older is a Girl BG, GG Two girls given older is a girl GG Slide 12-63 Example Use the results of the taste test given at a local mall. If one person from the sample is selected at random, find the probability the person selected Prefers Peppermint Prefers Wintergreen Total Men 70 55 125 Women 60 72 132 Total 130 127 257 Copyright © 2005 Pearson Education, Inc. Slide 12-64 Example continued a) prefers peppermint 130 P (peppermint) 257 b) is a woman 132 P (woman) 257 Copyright © 2005 Pearson Education, Inc. Slide 12-65 Example continued c) prefers peppermint, given that a woman is selected 60 15 P (peppermint|woman) 132 33 d) is a man, given that the person prefers wintergreen 55 P (man|wintergreen) 127 Copyright © 2005 Pearson Education, Inc. Slide 12-66 12.8 The Counting Principle and Permutations Copyright © 2005 Pearson Education, Inc. Counting Principle If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct ways, then the two experiments in that specific order can be performed in M • N ways. Copyright © 2005 Pearson Education, Inc. Slide 12-68 Example A password to logon to a computer system is to consist of 3 letters followed by 3 digits. Determine how many different passwords are possible if a) repetition of letters and digits is permitted b) repetition of letters and digits is not permitted c) the first letter must be a vowel (a, e, I, o, u) and the first digit cannot be 0, and repetition of letters and digits is not permitted. Copyright © 2005 Pearson Education, Inc. Slide 12-69 Solutions repetition of letters and digits is permitted. There are 26 letters and 10 digits. We have 6 positions to fill. L L L D D D 26 26 26 10 10 10 L L L D D D 17,576,000 Copyright © 2005 Pearson Education, Inc. Slide 12-70 Solution repetition of letters and digits is not permitted. L L L D D D 26 25 24 10 9 L L L 8 D D D 11,232,000 Copyright © 2005 Pearson Education, Inc. Slide 12-71 Solution the first letter must be a vowel (a, e, i, o, u) and the first digit cannot be 0, and repetition of letters and digits is not permitted. L L L D D D 5 25 24 9 L D D D L L 9 8 1,944,000 Copyright © 2005 Pearson Education, Inc. Slide 12-72 Permutations A permutation is any ordered arrangement of a given set of objects. (Order is important) Number of Permutations The number of permutations of n distinct items is n factorial, symbolized n!, where n! = n(n 1)(n 2) … (3)(2)(1) Copyright © 2005 Pearson Education, Inc. Slide 12-73 Example How many different ways can 6 stuffed animals be arranged in a line on a shelf? 6! = 6 • 5 • 4 • 3 • 2 • 1 = 720 The 6 stuffed animals can be arranged in 720 different ways. Copyright © 2005 Pearson Education, Inc. Slide 12-74 Example Consider the six numbers 1, 2, 3, 4, 5 and 6. In how many distinct ways can three numbers be selected and arranged if repetition is not allowed? 6 • 5 • 4 = 120 Thus, there are 120 different possible ordered arrangements, or permutations. Copyright © 2005 Pearson Education, Inc. Slide 12-75 Permutation Formula The number of permutations possible when r objects are selected from n objects is found by the permutation formula n! n Pr n r ! Copyright © 2005 Pearson Education, Inc. Slide 12-76 Example The swimming coach has 8 swimmers who can compete in the 100 m butterfly, he can only enter 3 swimmers in the event. In how many ways could he select the 3 swimmers? 8! 8! 8 P3 8 3 ! 5! 8 7 6 5 4 3 2 1 5 4 3 2 1 336 Copyright © 2005 Pearson Education, Inc. Slide 12-77 Permutations of Duplicate Objects The number of distinct permutations of n objects where n1 of the objects are identical, n2 of the objects are identical, …, nr of the objects are identical is found by the formula n! n1 ! n2 !...n3 ! Copyright © 2005 Pearson Education, Inc. Slide 12-78 Example In how many different ways can the letters of the word “CINCINNATI” be arranged? Of the 10 letters, 2 are C’s, 3 are N’s, and 3 are I’s. 10! 10 9 8 7 6 5 4 3 2 1 3!3!2! 3 2 1 3 2 1 2 1 Copyright © 2005 Pearson Education, Inc. 100,800 50,400 2 Slide 12-79 12.9 Combinations Copyright © 2005 Pearson Education, Inc. Combination A combination is a distinct group (or set) of objects without regard to their arrangement. The number of combinations possible when r objects are selected from n objects is found by n! n Cr n r ! r ! Copyright © 2005 Pearson Education, Inc. Slide 12-81 Example A student must select 4 of 7 essay questions to be answered on a test. In how many ways can this selection be made? 7! 7! 7 C4 (7 4)!4! 3!4! 7 6 5 4 3 2 1 35 3 2 1 4 3 2 1 There are 35 different ways that 4 of 7 questions can be selected. Copyright © 2005 Pearson Education, Inc. Slide 12-82 Example Toastline Bakery is testing 5 new wheat breads, 4 bran breads and 3 oat breads. If it plans to market 2 of the wheat breads, 2 of the bran breads and one of the oat breads, how many different combinations are possible? Bread choices 5C2 4C2 3 C1 10 6 3 180 Copyright © 2005 Pearson Education, Inc. Slide 12-83 12.10 Solving Probability Problems by Using Combinations Copyright © 2005 Pearson Education, Inc. Example A club consists of 5 men and 6 women. Four members are to be selected at random to form a committee. What is the probability that the committee will consist of two women? P (two women) = # of possible committees with two women total number of 4-member committees 6 C2 11C4 15 1 330 22 Copyright © 2005 Pearson Education, Inc. Slide 12-85 Example The Honey Bear is testing 10 new flavors of ice cream. They are testing 5 vanilla based, 3 chocolate based and 2 strawberry based ice creams. If we assume that each of the 10 flavors has the same chance of being selected and that 4 new flavors will be produced, find the probability that a) no chocolate flavors are selected. b) at least 1 chocolate is selected. c) 2 vanilla and 2 chocolate are selected. Copyright © 2005 Pearson Education, Inc. Slide 12-86 Solution 5 vanilla, 3 chocolate, 2 strawberry selecting 4 flavors a) P(no chocolate) 7C4 35 1 10 C4 210 6 b) P (at least 1 chocolate) 1 P (no chocolate) 1 Copyright © 2005 Pearson Education, Inc. 1 5 6 6 Slide 12-87 Solution continued C2 3 C2 P (2 vanilla and 2 chocolate) 10 C4 5 Copyright © 2005 Pearson Education, Inc. 10 3 30 1 210 210 7 Slide 12-88 12.11 Binomial Probability Formula Copyright © 2005 Pearson Education, Inc. To Use the Binomial Probability Formula There are n repeated independent trials. Each trial has two possible outcomes, success and failure. For each trial, the probability of success (and failure) remains the same. Copyright © 2005 Pearson Education, Inc. Slide 12-90 Binomial Probability Formula The probability of obtaining exactly x successes, P(x), in n independent trials is given by x nx P( x ) nCx p q where p is the probability of success on a single trial and q is the probability of failure on a single trial. Copyright © 2005 Pearson Education, Inc. Slide 12-91 Example A basket contains 5 pens: one of each color red, blue, black, green and purple. Five pens are going to be selected with replacement from the basket. Find the probability that no blue pens are selected exactly 1 blue pen is selected exactly 2 blue pens are selected exactly 3 blue pens are selected Copyright © 2005 Pearson Education, Inc. Slide 12-92 Solution no blue pens P ( x ) n Cx p x q n x 0 1 4 P (0) 5 C0 5 5 4 (1)(1) 5 5 5 1024 4 3125 5 Copyright © 2005 Pearson Education, Inc. exactly 1 blue pen P ( x ) n Cx p x q n x 5 0 1 1 4 P (0) 5 C1 5 5 5 1 4 1 4 (5) 5 5 1 256 256 5 5 625 625 Slide 12-93 Solution continued exactly 2 blue pens P ( x ) n Cx p x q n x P ( x ) n Cx p x q n x 2 1 4 P (0) 5 C2 5 5 2 3 52 1 4 (10) 5 5 1 64 128 10 25 125 625 Copyright © 2005 Pearson Education, Inc. exactly 3 blue pens 3 1 4 P (0) 5 C3 5 5 3 5 3 2 1 4 (5) 5 5 1 16 32 10 125 25 625 Slide 12-94 Example A manufacturer of lunch boxes knows that 0.7% of the lunch boxes are defective. Write the binomial probability formula that would be use to determine the probability that exactly x our of n lunch boxes produced are defective. Write the binomial probability formula that would be used to find the probability that exactly 4 lunch boxes out of 60 produced will be defective. Copyright © 2005 Pearson Education, Inc. Slide 12-95 Solution P ( x ) n Cx p x q n x n Cx (0.007)x ((0.993)n x P ( x ) n Cx p x q n x 60 C4 (0.007) ((0.993) 4 Copyright © 2005 Pearson Education, Inc. 56 Slide 12-96 Example The probability that an egg selected at random has a weak shell and will crack is 0.2. Find the probability that none of 8 eggs selected at random has a weak shell. at least one of 8 eggs selected has a weak shell. Copyright © 2005 Pearson Education, Inc. Slide 12-97 Solution p = 0.2 q = 1 0.2 = 0.8 We want 0 successes in 8 trials. P ( x ) n Cx p x q n x 8 C0 (0.2)0 (0.8)8 0 1(1)(0.167772) 0.167772 at least 1egg = 1 0.167772 = 0.832228 Copyright © 2005 Pearson Education, Inc. Slide 12-98