Here’s another example that I’ve been asked in interviews. Your friend has two children. You don’t know the relative ages or even genders, this is all you know. You go to his house and knock on the door. A child answers (you can assume with probability = 1 that this is one of his two children.) Given that the child is a boy, what is the probability that the other is a girl? Before you observe the child, you would answer “1/2”, but now that you have information you need to update your probability. Possible cases (order matters, and the children are ordered by age): your friend could have a younger boy, older girl, two boys, two girls, or younger girl, older boy. So the event space is (b,g), (b,b), (g,g), (g,b). Before observing the boy, the probability that ONE of his children is a girl is 2/4 as seen above. But now that you observe the boy, there are now only three possibilities: (g,b), (b,b), (b,g) (can’t have (g,g).) The probability is 2/3. To use Bayes’ Theorem, let X = event of the other child being a girl and let Y be event that you have observed a boy. What is P(X|Y)? It’s P(X and Y)/P(Y). P(Y) = 3/4 (remember this is PRIOR probability!), P(girl and boy) = 1/2, (1/2)/3/4 = 2/3. Hope this helps…
yeap, exam was really !!! tough! i feel schweser is not enough! i’ve read and solved schweser 3 times ==>still failed
Good stuff DD.