1. 1 Learning Objectives Bayes’ Formula The student will be able to solve problems involving finding the probability of an earlier event conditioned on the occurrence of a later event using Bayes’ Formula. The student will be able to solve problems of the above type using a probability tree.

2. 2 Probability of an Earlier Event Given a Later Event A survey of middle-aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next 10 years. Men who are not balding in this way have an 11% probability of a heart attack. If a middle-aged man is randomly chosen, what is the probability that he is balding, given that he suffered a heart attack? See tree diagram on next slide.

3. 3 Tree Diagram We want P(B|H) = probability that he is balding, given that he suffered a heart attack. Balding 0.28 Not balding 0.72 heart attack 0.18 heart attack 0.11 no heart attack 0.82 no heart attack 0.89

4. 4 Derivation of Bayes’ Formula

5. 5 Solution of Problem = 0.389

6. 6 Another Method of Solution Balding 0.28 Not balding 0.72 heart attack 0.18 heart attack 0.11 no heart attack 0.82 no heart attack 0.89 (0.28)(0.18) =0.0504 (0.72)(0.11) =0.0792 Another way to look at this problem is that we know the person had a heart attack, so he has followed one of the two red paths. The sample space has been reduced to these paths. So the probability that he is balding is which is 0.0504/(0.0504 + 0.0792) = 0.389

7. 7 Summary of Tree Method of Solution You do not need to memorize Bayes’ formula. In practice, it is usually easier to draw a probability tree and use the following: Let U 1, U 2,…U n be n mutually exclusive events whose union is the sample space S. Let E be an arbitrary event in S such that P(E)  0. Then product of branch probabilities leading to E through U 1 P(U 1 |E) = sum of all branch products leading to E Similar results hold for U 2, U 3,…U n

8. Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? A well-known statement of the problem was published in Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990 (vos Savant 1990):Marilyn vos SavantParadevos Savant 1990 Vos Savant's response was that the contestant should always switch to the other door. If the car is initially equally likely to be behind each door, a player who picks Door 1 and doesn't switch has a 1 in 3 chance of winning the car while a player who picks Door 1 and does switch has a 2 in 3 chance. The host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car. Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong. (Tierney 1991) Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy.PhDsTierney 1991

9. Vos Savant's solution The solution presented by vos Savant in Parade (vos Savant 1990b) shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:vos Savant 1990b A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

10. Car is behind door # P(C=c)Player selects door # P(S=s)Host opens door # P(H=h)P(branch)Stay W=car L=goat Total P(W) SwitchTotal P(W) C=11/3 S=11/3 H=21/21/18W 1/18+ 1/18+ 1/18+ 1/18+ 1/18+ 1/18= 6/18 L 1/9+ 1/9+ 1/9+ 1/9+ 1/9+ 1/9= 6/9 H=31//21/18WL S=21/3H=311/9LW S=31/3H=211/9LW C=21/3 S=11/3H=311/9LW S=21/3 H=11/21/18WL H=31//21/18WL S=31//3H=111/9LW C=31/3 S=11/3H=211/9LW S=21/3H=111/9LW S=31/3 H=11/21/18WL H=21/21/18WL

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