11.6 Conditional Probability 715 RECREATIONAL MATH The Monty Hall Problem The Monty Hall problem is a probability problem based on the television game show Let’s Make a Deal and is named after the original host, Monty Hall (1921–2017). The problem became famous in 1990 in Marilyn vos Savant’s Ask Marilyn column in Parade magazine. The problem can be stated as follows. You are a contestant on Let’s Make a Deal ,’ and there are three doors to choose from: A B, , and C. Behind one door is a new car, and behind two doors are goats. You choose door A. Monty then reveals that behind door B is a goat. He then asks if you’d like to stay with your original choice, door A, or change your choice to door C. What should you do, stay with your original choice or change your choice? The calculations of the probabilities involved in this problem involve a formula called Bayes’ theorem . Although the calculation of these probabilities is beyond our level of discussion, there are several websites and smartphone apps that can simulate the Monty Hall problem. It can be shown, using Bayes’ theorem, that the probability of winning the new car if you stay with your original choice is . 1 3 However, the probability of winning the new car if you change your choice is ! 2 3 While these probabilities may be surprising, long term repeated trials of the problem using computer simulations confirm these results. We will discuss the Monty Hall problem and Bayes’ theorem again in Exercise 84. m Monty Hall Here, the number of elements in E1 is five, the number of elements in E2 is six, and the number of elements in both E1 and E ,2 or E E , 1 2 ∩ is two. = = P E E n E E n E ( ) ( and ) ( ) 2 5 2 1 1 2 1 Thus, for this situation, the probability of selecting an element from E ,2 given that the element is in E ,1 is . 2 5 Example 3 Using the Conditional Probability Formula Two hundred and fifty patients who had knee, hip, or heart surgery were asked whether they were satisfied or dissatisfied regarding the results of their surgery. The responses are given in the table below. Surgery Satisfied Dissatisfied Total Knee 75 20 95 Hip 90 15 105 Heart 45 5 50 Total 210 40 250 If one person from the 250 patients surveyed is randomly selected, determine the probability that the person a) was satisfied with the results of the surgery. b) was satisfied with the results of the surgery, given that the person had knee surgery. c) was dissatisfied with the results of the surgery, given that the person had hip surgery. d) had heart surgery, given that the person was dissatisfied with the results of the surgery. Solution a) The total number of patients is 250, of which 210 were satisfied with the results of the surgery. P(satisfied with the results of the surgery) 210 250 21 25 = = b) We are given that the person had knee surgery. Thus, we have a conditional probability problem. Let E1 be the given information “the person had knee surgery.” Let E2 be “the person was satisfied with the results of the surgery.” We are being asked to determine P E E ( ). 2 1 The number of people who had knee surgery, n E( ), 1 is 95. The number of people who had knee surgery and were satisfied with the results of the surgery, n E E ( and ), 1 2 is 75. Thus, = = = P E E n E E n E ( ) ( and ) ( ) 75 95 15 19 2 1 1 2 1 c) We are given that the person had hip surgery. Thus, we have a conditional probability problem. Let E1 be the given information “the person had hip surgery.” Let E2 be “the person was dissatisfied with the results of the surgery.” We are asked to determine P E E ( ). 2 1 The number of people who had hip S_bukley/Shuttersock
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