A friend who works in a big city owns two cars, one small and one large. One-quarter of the time he drives the small car to work, and three-quarters of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.8. If he takes the large car, he is on time to work with probability 0.5. Given that he was at work on time on a particular morning, what is the probability that he drove the large car?
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Let S denote the event he takes the small car.
P(S) = 0.25 and P(~S) = 0.75
Let X denote the event he is on time for work.
P(X|S) = 0.8 and P(X|~S) = 0.5
P(S and X) = P(X|S)P(S) = 0.8*0.25 = 0.2
P(~S and X) = P(X|~S)P(~S) = 0.5*0.75 = 0.375
P(X) = P(X and S) + P(X and ~S) = 0.2 + 0.375 = 0.575
P(~S|X) = P(~S and X) / P(X) = 0.375 / 0.575 = 15/23
That's 0.6522 to 4 d.p.
P(S) = 0.25 and P(~S) = 0.75
Let X denote the event he is on time for work.
P(X|S) = 0.8 and P(X|~S) = 0.5
P(S and X) = P(X|S)P(S) = 0.8*0.25 = 0.2
P(~S and X) = P(X|~S)P(~S) = 0.5*0.75 = 0.375
P(X) = P(X and S) + P(X and ~S) = 0.2 + 0.375 = 0.575
P(~S|X) = P(~S and X) / P(X) = 0.375 / 0.575 = 15/23
That's 0.6522 to 4 d.p.