Chapter 15, \"More Probability\" from Applied Finite Mathematics by Rupinder Sekhon was developed by OpenStax College, licensed by Rice University, and is available on the Connexions website. It is used under a Creative Commons Attribution 3.0 Unported license. Chapter 15 1 More Probability 15.1 Chapter Overview In this chapter, you will learn to: 1. 2. 3. 4. Find Find Find Find the probability of a binomial experiment. probabilities using Bayes' Formula. the expected value or payo\u001b in a game of chance. probabilities using tree diagrams. 15.2 Binomial Probability In this section, we will consider types of problems that involve a sequence of trials, where each trial has only two outcomes, a success or a failure. These trials are independent, that is, the outcome of one does not a\u001bect the outcome of any other trial. Furthermore, the probability of success, p, and the probability of failure, (1 p), remains the same throughout the experiment. These problems are called binomial probability problems. Since these problems were researched by a Swiss mathematician named Jacques Bernoulli around 1700, they are also referred to as Bernoulli trials. We give the following de\u001cnition: Binomial Experiment A binomial experiment satis\u001ces the following four conditions: 1. There are only two outcomes, a success or a failure, for each trial. 2. The same experiment is repeated several times. 3. The trials are independent; that is, the outcome of a particular trial does not a\u001bect the outcome of any other trial. 4. The probability of success remains the same for every trial. The probability model that we are about to investigate will give us the tools to solve many real-life problems like the ones given below. 1. If a coin is \u001dipped 10 times, what is the probability that it will fall heads 3 times? 2. If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 7 out of 10 free throws in a game? 3. If a medicine cures 80% of the people who take it, what is the probability that among the ten people who take the medicine, 6 will be cured? 1 This content is available online at . Available for free at Connexions 251 252 CHAPTER 15. MORE PROBABILITY 4. If a microchip manufacturer claims that only 4% of his chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective? 5. If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product? We now consider the following example to develop a formula for \u001cnding the probability of k successes in n Bernoulli trials. Example 15.1 A baseball player has a batting average of .300. If he bats four times in a game, \u001cnd the probability that he will have a. b. c. d. e. four hits three hits two hits one hit no hits. Solution Let us suppose S denotes that the player gets a hit, and F denotes that he does not get a hit. This is a binomial experiment because it meets all four conditions. First, there are only two outcomes, S or F . Clearly the experiment is repeated four times. Lastly, if we assume that the player's skillfulness to get a hit does not change each time he comes to bat, the trials are independent with a probability of .3 of getting a hit during each trial. We draw a tree diagram to show all situations. Available for free at Connexions 253 Figure 15.1 Let us \u001crst \u001cnd the probability of getting, for example, two hits. We will have to consider the six possibilities, SSFF, SFSF, SFFS, FSSF, FSFS, FFSS, as shown in the above tree diagram. We list the probabilities of each below. 2 2 P (SSFF) = (.3) (.3) (.7) (.7) = (.3) (.7) 2 2 P (SFSF) = (.3) (.7) (.3) (.7) = (.3) (.7) 2 2 P (SFFS) = (.3) (.7) (.7) (.3) = (.3) (.7) 2 2 P (FSSF) = (.7) (.3) (.3) (.7) = (.3) (.7) 2 2 P (FSFS) = (.7) (.3) (.7) (.3) = (.3) (.7) Available for free at Connexions 254 CHAPTER 15. 2 MORE PROBABILITY 2 P (FFSS) = (.7) (.7) (.3) (.3) = (.3) (.7) 2 2 Since the probability of each of these six outcomes is (.3) (.7) , the probability of obtaining two 2 2 successes is 6(.3) (.7) . The probability of getting one hit can be obtained in the same way. Since each permutation has one S and three F 's, there are four such outcomes: SFFF, FSFF, FFSF, and FFFS. 3 And since the probability of each of the four outcomes is (.3) (.7) , the probability of getting 3 one hit is 4 (.3) (.7) . The table below lists the probabilities for all cases, and shows a comparison with the binomial expansion of fourth degree. Again, p denotes the probability of success, and q = (1 p) the probability of failure. Outcome Probability Four Hits (.3) 4 Three hits 3 4(.3) (.7) Two Hits 2 6(.3) (.7) One hits 2 4 (.3) (.7) No Hits 3 (.7) 4 Table 15.1 This gives us the following theorem: Theorem 15.1: Binomial Probability Theorem The probability of obtaining k successes in n independent Bernoulli trials is given by P (n, k; p) = nCkpk q nk (15.1) where p denotes the probability of success and q = (1 p)the probability of failure. We use the above formula to solve the following examples. Example 15.2 If a coin is \u001dipped 10 times, what is the probability that it will fall heads 3 times? Solution Let S denote the probability of obtaining a head, and F the probability of obtaining a tail. Clearly, n = 10, k = 3, p = 1/2, and q = 1/2. Therefore, 3 7 b (10, 3; 1/2) = 10C3(1/2) (1/2) = .1172 (15.2) Example 15.3 If a basketball player makes 3 out of every 4 free throws, what is the probability that he will make 6 out of 10 free throws in a game? Solution The probability of making a free throw is 3/4. Therefore, p = 3/4, q = 1/4, n = 10, and k = 6. Therefore, 6 4 b (10, 6; 3/4) = 10C6(3/4) (1/4) = .1460 Available for free at Connexions (15.3) 255 Example 15.4 If a medicine cures 80% of the people who take it, what is the probability that of the eight people who take the medicine, 5 will be cured? Solution Here p = .80, q = .20, n = 8, and k = 5. 5 3 b (8, 5; .80) = 8C5(.80) (.20) = .1468 (15.4) Example 15.5 If a microchip manufacturer claims that only 4% of his chips are defective, what is the probability that among the 60 chips chosen, exactly three are defective? Solution If S denotes the probability that the chip is defective, and F the probability that the chip is not defective, then p = .04, q = .96, n = 60, and k = 3. 3 b (60, 3; .04) = 60C3(.04) (.96) 57 = .2138 (15.5) Example 15.6 If a telemarketing executive has determined that 15% of the people contacted will purchase the product, what is the probability that among the 12 people who are contacted, 2 will buy the product? Solution If S denoted the probability that a person will buy the product, and F the probability that the person will not buy the product, then p = .15, q = .85, n = 12, and k = 2. 2 10 b (12, 2, .15) = 12C2(.15) (.85) = .2924. 15.3 Bayes' Formula In this section, we will develop and use Bayes' Formula to solve an important type of probability problem. Bayes' formula is a method of calculating the conditional probability P (F | E) from P (E | F ). The ideas involved here are not new, and most of these problems can be solved using a tree diagram. However, Bayes' formula does provide us with a tool with which we can solve these problems without a tree diagram. We begin with an example. Example 15.7 Suppose you are given two jars. Jar I contains one black and 4 white marbles, and Jar II contains 4 black and 6 white marbles. If a jar is selected at random and a marble is chosen, Available for free at Connexions 256 CHAPTER 15. MORE PROBABILITY a. What is the probability that the marble chosen is a black marble? b. If the chosen marble is black, what is the probability that it came from Jar I? c. If the chosen marble is black, what is the probability that it came from Jar II? Solution Let JI I be the event that Jar I is chosen, J II be the event that Jar II is chosen, B be the event that a black marble is chosen and W the event that a white marble is chosen. We illustrate using a tree diagram. (a) (b) Figure 15.2 a. The probability that a black marble is chosen is P (B) = 1/10 + 2/10 = 3/10. b. To \u001cnd P (JI | B), we use the de\u001cnition of conditional probability, and we get P (JI | B) = P (JI B) 1/10 1 = = P (B) 3/10 3 (15.6) IIB) 2/10 c. Similarly, P (J II | B) = P (J = 3/ = 23 P (B) 10 In parts b and c, the reader should note that the denominator is the sum of all probabilities of all branches of the tree that produce a black marble, while the numerator is the branch that is associated with the particular jar in question. We will soon discover that this is a statement of Bayes' formula . Let us \u001crst visualize the problem. We are given a sample space S and two mutually exclusive events JI and J II. That is, the two events, JI and J II, divide the sample space into two parts such that JI J II = S . Furthermore, we are given an event B that has elements in both JI and J II, as shown in the Venn diagram below. Available for free at Connexions 257 (a) (b) Figure 15.3 From the Venn diagram, we can see that B = (B JI) (B J II) and P (B) = P (B JI) + P (B J II) But the product rule in Chapter 13 gives us P (B JI) = P (JI) P (B | JI) P (B J II) = P (J II) P (B | J II) Substituting in p. 257, we get P (B) = P (JI) P (B | JI) + P (J II) P (B | J II) The conditional probability formula gives us P (JI | B) = P (JIB) P (B) Therefore, (B|JI)) P (JI | B) = P (JIP P (B) or, P (JI)P (B|JI) P (JI | B) = P (JI)P (B|JI)+P (J II)P (B|J II) The last statement is Bayes' Formula for the case where the sample space is divided into two partitions. The following is the generalization of this formula for n partitions. 15.8 Let S be a sample space that is divided into n partitions, A1 , A2 , . . . An . If E is any event in S , then P (Ai | E) = P (Ai ) P (E | Ai ) P (A1 ) P (E | A1 ) + P (A2 ) P (E | A2 ) + + P (An ) P (E | An ) (15.7) We begin with the following example. Example 15.9 A department store buys 50% of its appliances from Manufacturer A, 30% from Manufacturer B, and 20% from Manufacturer C. It is estimated that 6% of Manufacturer A's appliances, 5% of Manufacturer B's appliances, and 4% of Manufacturer C's appliances need repair before the warranty expires. An appliance is chosen at random. If the appliance chosen needed repair before Available for free at Connexions 258 CHAPTER 15. MORE PROBABILITY the warranty expired, what is the probability that the appliance was manufactured by Manufacturer A? Manufacturer B? Manufacturer C? Solution Let events A, B and C be the events that the appliance is manufactured by Manufacturer A, Manufacturer B, and Manufacturer C, respectively. Further, suppose that the event R denotes that the appliance needs repair before the warranty expires. We need to \u001cnd P (A | R), P (B | R) and P (C | R). We will do this problem both by using a tree diagram and by using Bayes' formula. We draw a tree diagram. Figure 15.4 The probability P (A | R), for example, is a fraction whose denominator is the sum of all probabilities of all branches of the tree that result in an appliance that needs repair before the warranty expires, and the numerator is the branch that is associated with Manufacturer A. P (B | R) and P (C | R) are found in the same way. We list both as follows: P (A | R) = (.030)+(.030 = .030 = .566 .015)+(.008) .053 .015 .008 P (B | R) = .053 = .283 and P (C | R) = .053 = .151. Alternatively, using Bayes' formula, P (A | R) = = P (A)P (R|A) P (A)P (R|A)+P (B)P (R|B)+P (C)P (R|C) .030 (.030)+(.015)+(.008) = .030 .053 = .566 Available for free at Connexions (15.8) 259 P (B | R) and P (C | R) can be determined in the same manner. Example 15.10 There are \u001cve Jacy's department stores in San Jose. The distribution of number of employees by gender is given in the table below. Store Number Number of Employees Percent of Women Employees 1 300 .40 2 150 .65 3 200 .60 4 250 .50 5 100 .70 Total=1000 Table 15.2 If an employee chosen at random is a woman, what is the probability that the employee works at store III? Solution Let k = 1, 2, ..., 5 be the event that the employee worked at store k , and W be the event that the employee is a woman. Since there are a total of 1000 employees at the \u001cve stores, P (1) = .30 P (2) = .15 P (3) = .20 P (4) = .25 P (5) = .10 (15.9) Using Bayes' formula, P (3, |, W ) = = P (3)P (W,|,3) P (1)P (W,|,1)+P (2)P (W,|,2)+P (3)P (W,|,3)+P (4)P (W,|,4)+P (5)P (W,|,5) (.,20)(.,60) (.,30)(.,40)+(.,15)(.,65)+(.,20)(.,60)+(.,25)(.,50)+(.,10)(.,70) (15.10) = .2254 15.4 Expected Value An expected gain or loss in a game of chance is called Expected Value. The concept of expected value is closely related to a weighted average. Consider the following situations. 1. Suppose you and your friend play a game that consists of rolling a die. Your friend o\u001bers you the following deal: If the die shows any number from 1 to 5, he will pay you the face value of the die in dollars, that is, if the die shows a 4, he will pay you $4. But if the die shows a 6, you will have to pay him $18. Before you play the game you decide to \u001cnd the expected value. You analyze as follows. Available for free at Connexions 260 CHAPTER 15. MORE PROBABILITY Since a die will show a number from 1 to 6, with an equal probability of 1/6, your chance of winning $1 is 1/6, winning $2 is 1/6, and so on up to the face value of 5. But if the die shows a 6, you will lose $18. You write the expected value. E = $1 (1/6) + $2 (1/6) + $3 (1/6) + $4 (1/6) + $5 (1/6) $18 (1/6) = $.50 This means that every time you play this game, you can expect to lose 50 cents. In other words, if you play this game 100 times, theoretically you will lose $50. Obviously, it is not to your interest to play. 2. Suppose of the ten quizzes you took in a course, on eight quizzes you scored 80, and on two you scored 90. You wish to \u001cnd the average of the ten quizzes. The average is 8 2 (80) (8) + (90) (2) = (80) + (90) = 82 (15.11) 10 10 10 It should be observed that it will be incorrect to take the average of 80 and 90 because you scored 80 on eight quizzes, and 90 on only two of them. Therefore, you take a "weighted average" of 80 and 90. That is, the average of 8 parts of 80 and 2 parts of 90, which is 82. A= In the \u001crst situation, to \u001cnd the expected value, we multiplied each payo\u001b by the probability of its occurrence, and then added up the amounts calculated for all possible cases. In the second part of p. 259, if we consider our test score a payo\u001b, we did the same. This leads us to the following de\u001cnition. De\u001cnition 15.1: Expected Value If an experiment has the following probability distribution, Payo\u001b x1 Probability p (x1 ) p (x2 ) p (x3 ) x2 x3 xn p (xn ) Table 15.3 then the expected value of the experiment is Expected Value = x1 p (x1 ) + x2 p (x2 ) + x3 p (x3 ) + + xn p (xn ) Example 15.11 In a town, 10% of the families have three children, 60% of the families have two children, 20% of the families have one child, and 10% of the families have no children. What is the expected number of children to a family? Solution We list the information in the following table. Number of Children 3 2 1 0 Probability .10 .60 .20 .10 Table 15.4 Expected Value = x1 p (x1 ) + x2 p (x2 ) + x3 p (x3 ) + x4 p (x4 ) (15.12) E = 3 (.10) + 2 (.60) + 1 (.20) + 0 (.10) = 1.7 (15.13) So on average, there are 1.7 children to a family. Available for free at Connexions 261 Example 15.12 To sell an average house, a real estate broker spends $1200 for advertisement expenses. If the house sells in three months, the broker makes $8,000. Otherwise, the broker loses the listing. If there is a 40% chance that the house will sell in three months, what is the expected payo\u001b for the real estate broker? Solution The broker makes $8,000 with a probability of .40, but he loses $1200 whether the house sells or not. E = ($.8000) (.40) ($1200) = $2, 000. Alternatively, the broker makes $ (8000 1200) with a probability of .40, but loses $1200 with a probability of .60. Therefore, E = ($6800) (.40) ($1200) (.60) = $2, 000. Example 15.13 In a town, the attendance at a football game depends on the weather. On a sunny day the attendance is 60,000, on a cold day the attendance is 40,000, and on a stormy day the attendance is 30,000. If for the next football season, the weatherman has predicted that 30% of the days will be sunny, 50% of the days will be cold, and 20% days will be stormy, what is the expected attendance for a single game? Solution Using the expected value formula, we get e = (60, 000) (.30) + (40, 000) (.50) + (30, 000) (.20) = 44, 000. (15.14) Example 15.14 A lottery consists of choosing 6 numbers from a total of 51 numbers. The person who matches all six numbers wins $2 million. If the lottery ticket costs $1, what is the expected payo\u001b? Solution Since there are 51C6 = 18, 009, 460 combinations of six numbers from a total of 51 numbers, the chance of choosing the winning number is 1 out of 18,009,460. So the expected payo\u001b is \u0013 \u0012 1 $1 = $0.89 (15.15) E = ($2 million) 18009460 This means that every time a person spends $1 to buy a ticket, he or she can expect to lose 89 cents. 15.5 Probability Using Tree Diagrams As we have already seen, tree diagrams play an important role in solving probability problems. A tree diagram helps us not only visualize, but also list all possible outcomes in a systematic fashion. Furthermore, when we list various outcomes of an experiment and their corresponding probabilities on a tree diagram, we Available for free at Connexions 262 CHAPTER 15. MORE PROBABILITY gain a better understanding of when probabilities are multiplied and when they are added. The meanings of the words and and or become clear when we learn to multiply probabilities horizontally across branches, and add probabilities vertically down the tree. Although tree diagrams are not practical in situations where the possible outcomes become large, they are a signi\u001ccant tool in breaking the problem down in a schematic way. We consider some examples that may seem di\u001ecult at \u001crst, but with the help of a tree diagram, they can easily be solved. Example 15.15 A person has four keys and only one key \u001cts to the lock of a door. What is the probability that the locked door can be unlocked in at most three tries? Solution Let U be the event that the door has been unlocked and L be the event that the door has not been unlocked. We illustrate with a tree diagram. Figure 15.5 The probability of unlocking the door in the \u001crst try = 1/4 Available for free at Connexions (15.16) 263 The probability of unlocking the door in the second try = (3/4) (1/3) = 1/4 The probability of unlocking the door in the third try = (3/4) (2/3) (1/2) = 1/4 (15.17) (15.18) Therefore, the probability of unlocking the door in at most three tries = 1/4 + 1/4 + 1/4 = 3/4 Example 15.16 A jar contains 3 black and 2 white marbles. We continue to draw marbles one at a time until two black marbles are drawn. If a white marble is drawn, the outcome is recorded and the marble is put back in the jar before drawing the next marble. What is the probability that we will get exactly two black marbles in at most three tries? Solution We illustrate using a tree diagram. Figure 15.6 The probability that we will get two black marbles in the \u001crst two tries is listed adjacent to the 3 lowest branch, and it = 10 3 The probability of getting \u001crst black, second white, and third black = 20 3 Similarly, the probability of getting \u001crst white, second black, and third black = 25 3 Therefore, the probability of getting exactly two black marbles in at most three tries = 10 += 3 3 57 + = = 20 25 100 Available for free at Connexions 264 CHAPTER 15. MORE PROBABILITY Example 15.17 A circuit consists of three resistors: resistor R1 , resistor R2 , and resistor R3 , joined in a series. If one of the resistors fails, the circuit stops working. If the probability that resistors R1 , R2 , or R3 will fail is .07, .10, and .08, respectively, what is the probability that at least one of the resistors will fail? Solution Clearly, the that at least one of the resistors fails = 1 none of the resistors fails. It is quite easy to \u001cnd the probability of the event that none of the resistors fails. We don't even need to draw a tree because we can visualize the only branch of the tree that assures this outcome. The probabilities that R1 , R2 , R3 will not fail are .93, .90, and .92 respectively. Therefore, the probability that none of the resistors fails = (.93) (.90) (.92) = .77. Thus, the probability that at least one of them will fail = 1 .77 = .23. Available for free at Connexions