The Two Children Problem

If 张三 has two children, and one of whom is a boy, what's the probability that he has two boys?


Clearly the answer isn't 1/2, otherwise I wouldn't be asking. Indeed, the two events are independent. The hinge lies on the first event being ambiguous: we're not told which of the two children is the boy; it's not fixed. Thus, the first event (one of the children is a boy) is actually two: child A is, in which case B is the unknown, or child B is the boy and A is the unknown.


Tabulating all of the possibilities helps:
     A | B
     ____

1.  g  |  g // ignore this one

2.  b  |  g

3.  g  |  b

4.  b  |  b

We can eliminate 1., because it's given that A or B is a boy. Thus, we're left with two cases of one boy and one case of two boys. Hence, 1/3.


Or, with conditional probability:
Let A be the event child A is a boy.
Let B be the event child B is a boy.
Let C be the event that P(A U B) = 1. //given
//P(P | Q) means "the probability of P, given Q."
//P(P^Q) means "the probability of P and Q both happening)."


P((P^Q) | C) = P(A^B^C) / P(C)  
                 = P(A^B) / (3/4)  
                 = 4(.25) / 3 = 1/3         //A and B are independent
You can drop the C from P(A^B^C) because for A^B to occur, C also has to occur. For example:
Let M be the event that you roll a die.
Let N be the event that you roll a six.
P(M^N) = P(N) . Or, more specifically, P(N^M') = 0, where M' means "not M." In English, the probability of you rolling a six and you don't roll a die both happening is zero.