Not what you think

We know from Bayes' Theorem that the probability that there are 2 boys, given one is born on Tuesday is

$P(BB|B_T) = \dfrac{P(B_T|BB)P(BB)}{P(B_T)}$

We will find a general solution to this first. Let's assume that the probability of a boy being born on a Tuesday is $\epsilon$ (we will substitute $\epsilon = \frac{1}{7}$ later).

Thus, $P(B_T|BB)$ is the probability of at least one boy being born on a Tuesday, given that the family has two boys, which is equivalent to $1-(1-\epsilon)^2$ (one minus the probability that neither boys are born on Tuesday).

$P(BB)$ (probability of two boys) is just $\frac{1}{4}$ .

$P(B_T)$ (probability of a boy being born on a Tuesday) is trickier, since we have to consider the 4 sample cases: $GG$ (two girls), $GB$ and $BG$ (a girl and a boy), and $BB$ (two boys). If it is $P(B_T(GG))$ , its value is 0, since there are no boys. If it's $P(B_T(GB))$ or $P(B_T(BG))$ , its value is $\epsilon$ since there is only one boy, whose chance of being born on Tuesday would be $\epsilon$ . If it is $P(B_T(BB))$ its value is (from PIE) $\epsilon+\epsilon-\epsilon^2$ since either one boy could be born on Tuesday, or the other, minus when both.

Thus, we have the equation as

$\begin{aligned} P(BB|B_T) &= \dfrac {(1-(1-\epsilon)^2)(\frac{1}{4})}{\frac{1}{4}(0+\epsilon+\epsilon+(\epsilon+\epsilon-\epsilon^2))}\\ &= \dfrac{1-(1-\epsilon)^2}{4\epsilon-\epsilon^2} \end{aligned}$

Substituting $\epsilon = \frac{1}{7}$ , we get

$\begin{aligned} P(BB|B_T) &= \dfrac {\frac{13}{49}}{\frac{27}{49}}\\ &= \dfrac {13}{27} \end{aligned}$

Therefore, the probability of 2 boys is $\frac{13}{27}$ , so the answer is $13+27=40$ .

One could also solve this by counting:

There are 14 equally likely cases in which the first child is a boy born on a Tuesday (In 7 of these, the other kid is a boy).

There are 14 equally likely cases in which the second child is a boy born on a Tuesday. (Again, in 7 of these the other kid is a boy).

But we have counted one case twice (In which both kids are boys born on Tuesdays). So we have a total of 14+14-1 = 27 equally likely cases in which one kid is a boy born on a Tuesday. And in 7+7-1 = 13 of these, the other kid is a boy. Hence the answer is 13/27.

The $14\times 14 = 196$ possible situations shown in the table are equally likely.

The shaded region corresponds with the knowledge that one of the children is a boy born on Tuesday. This leaves 27 possible situations.

The orange region corresponds with the situation that the other child is also a boy. It covers 13 possible situations.

Thus the conditional probability is $13/27$ , given the answer $\boxed{40}$ .

I had read about The Tuesday Birthday Problem Before , so I remembered the answer.