Question

I have two children. One is a boy born on a Tuesday. What is the probability that I have two boys?

Answer 1

Let's handle this case by case.

Clearly, there's no way both children can be girls.  There are then two cases:

Case 1: Two boys.  In this case, we have 13 possibilities:  the first is born on a Tuesday and the second is not (that's 6 possibilities, six ways to choose the day for the second boy), the first is not born on a Tuesday and the second is (6 more possibilities, same logic), and both are born on a Tuesday (1 final possibility), for a total of 13 possibilities with this case.

Case 2: A boy and a girl.  In this case, there are 14 possibilities:  The first is a boy born on a Tuesday and the second is a girl born on any day (7 possibilities, again choosing the day of the week.  We are counting possibilities by days of the week, so we must be consistent here.), or the first is a girl born any day and the second is a boy born on a Tuesday (7 possibilities).

We're trying to find the probability of case 1 occurring given that case 1 or case 2 occurs.  As there's 13+14=27 ways for either case to occur, we have a 13/27 probability that case 1 is the one that occurred.

Answer 2

Let B1T denote the first boy being born on Tuesday, B2N denote the second boy being born any day of the week and similarly B1N,B2T.

Then by Baye's formula, $P=\frac{P((B1T\cap B2N)\cup(B1N\cap B2T))_{B1\cap B2}P(B1\cap B2)}{P((B1T\cap B2N)\cup(B1N\cap B2T)}=\frac{(\frac{1}{7}+\frac{1}{7}-\frac{1}{7^2})\cdot\frac{1}4}{\frac{1}{14}+\frac{1}{14}-\frac{1}{14^2}}=\boxed{\frac{13}{27}}$