Chapter 8 Counting Principles; Further Probability Topics Copyright 2012 Pearson Education, Inc. All rights reserved 8.1 The Multiplication Principle; Permutations Copyright 2012 Pearson Education, Inc. All rights reserved 8- 3 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 A combination lock can be set to open to any 4-digit sequence. (a) How many sequences are possible? (b) How many sequences are possible if no digit is repeated? Solution: (a) Since there are 10 digits namely 0, 1, 2.....9, there are 10 choices for each of the digit. By the multiplication principle, there are 10 10 10 10 =10,000 different sequences. (b)There are 10 choices for the first digit. It cannot be used again, so there are 9 choices for the second digit, 8 choices for the third digit, and then 7 choices for the fourth digit. Consequently, the number of such sequences is 10 9 8 7 =5040 different sequences. 8- 4 2012 Pearson Education, Inc.. All rights reserved. Figure 1 8- 5 2012 Pearson Education, Inc.. All rights reserved. 8- 6 2012 Pearson Education, Inc.. All rights reserved. Figure 2 8- 7 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 A teacher is lining up 8 students for a spelling bee. How many different line-ups are possible? Solution: Eight choices will be made, one for each space that will hold a student. Any of the students could be chosen for the first space. There are 7 choices for the second space, since 1 student has already been placed in the first space; there are 6 choices for the third space, and so on. By the multiplication principle, the number of different possible arrangements is 8 7 6 5 4 3 2 1 = 40,320. 8- 8 2012 Pearson Education, Inc.. All rights reserved. 8- 9 2012 Pearson Education, Inc.. All rights reserved. 8 - 10 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 A teacher wishes to place 5 out of 8 different books on her shelf. How many arrangements of 5 books are possible? Solution : The teacher has 8 ways to fill the first space, 7 ways to fill the second space, 6 ways to fill the third, and so on... Since the teacher wants to use only 5 books, only 5 spaces can be filled (5 events) instead of 8, for 8 7 6 5 4 = 6720 arrangements. 8 - 11 2012 Pearson Education, Inc.. All rights reserved. 8 - 12 2012 Pearson Education, Inc.. All rights reserved. Your Turn 4 Find the number of permutations of the letters L, M, N, O, P, and Q, if just three of the letters are to be used. Solution: P(n, r ) n! . Here n 6 and r 3. (n - r )! 6! 6! (6 - 3)! 3! 6 3 21 5 4 120 3 2 1 8 - 13 2012 Pearson Education, Inc.. All rights reserved. Figure 3 8 - 14 2012 Pearson Education, Inc.. All rights reserved. Your Turn 6 In how many ways can the letters in the word Tennessee be arranged? Solution: If the n objects in a permutation are not all distinguishablethat is, if there are n1 of type 1, n2 of type 2, and so on for r different types, then the number of distinguishable permutations is n! . n1 !n2 ! ! nr This word contains 1 t, 4 e's, 2 n's, and 2 s's. To use the formula, let n = 9, n1 = 1, n2 = 4 , n3 = 2 and n4 = 2. 9! 1!4!2!2! = 3780 8 - 15 2012 Pearson Education, Inc.. All rights reserved. 8 - 16 2012 Pearson Education, Inc.. All rights reserved. 8.2 Combinations Copyright 2012 Pearson Education, Inc. All rights reserved 8 - 18 2012 Pearson Education, Inc.. All rights reserved. Figure 4 8 - 19 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 How many committees of 4 people can be formed from a group of 10 people? Solution: A committee is an unordered group, so use the combinations formula for C(10,4). 10! 10! C (10, 4) 4!(10 4)! 4!6! 10 7 6 5 3 21 10 9 7 9 8 4 8 210 4 16 5 3 21 4 1 3 2 4 3 2 8 - 20 2012 Pearson Education, Inc.. All rights reserved. 8 - 21 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 From a class of 15 students, a group of 3 or 4 students will be selected to work on a special project. In how many ways can a group of 3 or 4 students be selected? Solution: The number of ways to select group of 3 students from a class of 15 students is C(15, 3) = 455. The number of ways to select group of 4 students from a class of 15 students is C(15, 4) = 1365. The total number of ways to select a group of 3 or 4 students will be the 1820. 8 - 22 2012 Pearson Education, Inc.. All rights reserved. 8 - 23 2012 Pearson Education, Inc.. All rights reserved. 8 - 24 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 (a) How many 4-digit code numbers are possible if no digits are repeated? Solution: Since changing the order of the 4 digits results in a different code, permutations should be used. P (10, 4) 5040 (b) A sample of 3 light bulbs is randomly selected from a batch of 15. How many different samples are possible? Solution: The order in which the 3 light bulbs are selected is not important. The sample is unchanged if the items are rearranged, so combinations should be used. C (15,3) 455 8 - 25 2012 Pearson Education, Inc.. All rights reserved. Continued Your Turn 3 continued (c) In a baseball conference with 8 teams, how many games must be played so that each team plays every other team exactly once? Solution: Selection of 2 teams for a game is an unordered subset of 2 from the set of 8 teams. Use combinations again. C (8, 2) 28. (d) In how many ways can 4 patients be assigned to 6 different hospital rooms so that each patient has a private room? Solution: The room assignments are an ordered selection of 4 rooms from the 6 rooms. Exchanging the rooms of any 2 patients within a selection of 4 rooms gives a different assignment, so permutations should be used. P (6, 4) 360. 8 - 26 2012 Pearson Education, Inc.. All rights reserved. Figure 5 8 - 27 2012 Pearson Education, Inc.. All rights reserved. 8.3 Probability Applications of Counting Principles Copyright 2012 Pearson Education, Inc. All rights reserved Figure 6 8 - 29 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 What is the probability that 1 New York plant and 1 Chicago plant are selected? Solution: The probability for each branch is calculated by multiplying the probabilities. The resulting probabilities for the two branches are then added. 1 3 3 1 1 P (1 NY and 1 C) . 6 5 6 5 5 8 - 30 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 From a group of 22 nurses, if 8 of the 22 nurses are men, what is the probability that exactly 2 men are among the 4 nurses selected? Solution: Four nurses from a group of 22 can be selected in C (22, 4) ways. (Use combinations, since the group of 4 is an unordered set.) 22! (22)(21)(20)(19) C (22, 4) 7315. 18!4! (4)(3)(2)(1) There are C ( 8, 2) C (14, 2) ways to choose exactly 2 men among the 4 nurses. C (8, 2) (14, 2) (28) C (91) 2548. 8 - 31 2012 Pearson Education, Inc.. All rights reserved. Continued Your Turn 2 continued The probability that exactly 2 male nurses will be selected is given by n (E) / n (S), where E is the event that the chosen group includes exactly 2 male nurses, and S is the sample space for the experiment of choosing a group of 4 nurses. P exactly 2 male nurses n( E ) n( S ) 2548 7315 0.3483 8 - 32 2012 Pearson Education, Inc.. All rights reserved. Your Turn 3 When shipping diesel engines abroad, it is common to pack 12 engines in one container. Suppose that a company has received complaints from its customers that many of the engines arrive in nonworking condition. To help solve this problem, the company decides to make a spot check of containers after loading. The company will test 3 engines from a container at random; if any of the 3 are nonworking, the container will not be shipped until each engine in it is checked. Suppose a given container has 4 nonworking engines. Find the probability that the container will not be shipped. Solution: P (8,3) (0 C (4,0) (not shipping) = 1 P164defective in the sample) C (1)(56) 1 1 0.7455. C (12,3) 220 220 8 - 33 2012 Pearson Education, Inc.. All rights reserved. 8 - 34 2012 Pearson Education, Inc.. All rights reserved. Figure 7 8 - 35 2012 Pearson Education, Inc.. All rights reserved. 8.4 Binomial Probability Copyright 2012 Pearson Education, Inc. All rights reserved 8 - 37 2012 Pearson Education, Inc.. All rights reserved. 8 - 38 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Find the probability of getting exactly 4 heads in 8 tosses of a fair coin. Solution: The probability of success (getting a head in a single toss) is 1/2. The probability of a failure (getting a tail) is 1 1/ 2 = 1/ 2. Thus, 4 P(4 heads in 8 tosses) = C (8, 4) = 70 1 2 8 0.2734. 8 - 39 2012 Pearson Education, Inc.. All rights reserved. 1 1 2 2 4 Your Turn 3 Assuming that selection of items for a sample can be treated as independent trials, and that the probability that any 1 item is defective is 0.01, find the probability of 2 or 3 defective items in a random sample of 15 items from a production line. Solution: Here, a \"success\" is a defective item. Since selecting each item for the sample is assumed to be an independent trial, the binomial probability formula applies. The probability of success (a defective item) is 0.01, while the probability of failure (an acceptable item) is 0.99. This makes P(2 defective in 15 items) C (15, 2)(0.01) 2 (0.99)13 105(0.01) 2 (0.99)13 0.009214 8 - 40 2012 Pearson Education, Inc.. All rights reserved. Continued Your Turn 3 continued P(3 defective in 15 items) C (15,3)(0.01)3 (0.99)12 455(0.01)3 (0.99)12 0.000403 Use the union rule, noting that 2 defective and 3 defective are mutually exclusive events, to get P ( 2 defective or 3 defective items ) = P (2 defective items) + P (3 defective items). 0.009214 0.000403 0.009617 8 - 41 2012 Pearson Education, Inc.. All rights reserved. 8 - 42 2012 Pearson Education, Inc.. All rights reserved. 8.5 Probability Distributions; Expected Value Copyright 2012 Pearson Education, Inc. All rights reserved 8 - 44 2012 Pearson Education, Inc.. All rights reserved. 8 - 45 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 A shipment of 12 computer monitors contains 3 broken monitors. A shipping manager checks a sample of two monitors to see if any are broken. Find the probability distribution for the number of broken monitors. Solution: Let x represent the random variable \"number of broken monitors found by the manager.\" The possible values of x are 0, 1, or 2. There are 3 broken monitors, and 9 unbroken monitors, so the number of ways of choosing 0 broken monitors (which implies 2 unbroken monitors) is C (3,0) C (9, 2). The number of ways of choosing a sample of 2 monitors is C (12, 2). Continued 8 - 46 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued Therefore, the probability of choosing 0 broken monitors is C (3,0) (9, 2) (1) C (36) 6 P (0) . C (12, 2) 66 11 Therefore, the probability of choosing 1 broken monitor is C (3,1) (9,1) (3) C (9) 9 P (1) . C (12, 2) 66 22 Therefore, the probability of choosing 2 broken monitors is C (3, 2) (9,0) (3) C (1) 1 P(2) . C (12, 2) 66 22 Continued 8 - 47 2012 Pearson Education, Inc.. All rights reserved. Your Turn 1 continued Probability Distribution of Broken Monitors in Sample 8 - 48 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 Find the probability distribution and draw a histogram for the number of tails showing when three coins are tossed. Solution : Let x represent the random variable \"number of tails.\" Then x can take on the values 0, 1, 2, or 3. Now find the probability of each outcome. To find the probability of 0, 1, 2 or 3 tails, we can either use binomial probability, or notice that there are 8 outcomes in the sample space: {hhh, hht, hth, htt, thh, tht, tth, ttt}.The results are shown in the table with figure. Continued 8 - 49 2012 Pearson Education, Inc.. All rights reserved. Your Turn 2 continued Probability Distribution of Tails 8 - 50 2012 Pearson Education, Inc.. All rights reserved. Histogram Figure 8 8 - 51 2012 Pearson Education, Inc.. All rights reserved. Figure 9 8 - 52 2012 Pearson Education, Inc.. All rights reserved. Figure 10 8 - 53 2012 Pearson Education, Inc.. All rights reserved. 8 - 54 2012 Pearson Education, Inc.. All rights reserved. 8 - 55 2012 Pearson Education, Inc.. All rights reserved. 8 - 56 2012 Pearson Education, Inc.. All rights reserved. 8 - 57 2012 Pearson Education, Inc.. All rights reserved. 8 - 58 2012 Pearson Education, Inc.. All rights reserved. 8 - 59 2012 Pearson Education, Inc.. All rights reserved. Chapter 8 Review Extended Application: Optimal Inventory for a Service Truck Copyright 2012 Pearson Education, Inc. All rights reserved 8 - 61 2012 Pearson Education, Inc.. All rights reserved. ( ) A. C. B. D. A. B. C. Skip Navigation Homework Overview Name Unit VI Homework Last Worked 11/21/15 9:58pm Current Score 64.29% (22.50 points out of 35) This homework will affect your Study Plan score. Number of times you can complete each question: unlimited Reminder You must score at least 70% before you can begin Unit VI Assessment Changes made here WILL affect your score. Go to Results to practice without affecting your score. Questions: 35 Scored: 33 Correct: 21 Question 1 (1/1) Question 2 (1/1) Question 4 (1/1) Question 5 (1/1) Question 7 (1/1) Question 8 (1/1) Question 10 (0.50/1) Question 11 (1/1) Question 13 (1/1) Question 14 (0/1) Question 16 (1/1) Question 17 (1/1) Question 19 (0/1) Question 20 (0/1) Question 22 (1/1) Question 23 (1/1) Question 25 (0/1) Question 26 (0.50/1) Question 28 (0/1) Question 29 (1/1) Question 31 (1/1) Question 32 (0/1) Question 34 (0/1) Question 35 (0/1) Part To see what to study next, go to your Study Plan. This course is based on Lial: Finite Mathematics, 10e Copyright 2015 Pearson Education