Suppose a family has two kids of unknown gender and every time you visit, one of the kids will randomly answer the door. You go there on Monday and a boy answers. If you go back on Tuesday, what is the probability that a boy will answer?

I am solving the problem above and wanted to check if my solution is correct.

A: A boy answers on Monday
B: A boy answers on Tuesday

Possible kid's sex and the sex answering the door

1. BB : BB BB BB BB
2. BG : BB BG GB GG
3. GB : GG GB BG BB
4. GG : GG GG GG GG

The answer is asking $P(B|A)=$ $P(A \cap B)$ /$P(A)$

$P(A)=1/4\times1+1/4\times1/2+1/4\times1/2+1/4\times0$

$=1/4\times(1+1/2+1/2+0)=1/2$

$P(A \cap B)=1/4\times1+1/4\times1/4+1/4\times1/4+1/4\times0$

$=1/4\times(1+1/4+1/4+0)=3/8$

$P(B|A)=$ $P(A \cap B)$ /$P(A) = (3/8)/(1/2)=3/4$

Given multiple choices are 3/4, 2/3, 1/2, and 1/3.

I think it is 3/4 but I am not 100% sure. Could you see if my process is correct?

Thank you!