Suppose a family has two kids of unknown gender and every time you visit, one of the kids will randomly answer the door.

Question

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!

Answer 1

Assuming only $2$ genders, each equally probable, then we know one child is a boy and the other child is a boy with $50$% probability and a girl with $50$% probability. If there is no influence of previous days on the current day, then there is only a $25$% chance of a girl answering the door (product of probability of the other child answering the door and the probability the other child is girl). So, yes, $3/4$ chance a boy answers the door.

Answer 2

I came up with the same answer, via a somewhat different approach.

Initially either there are

$2$ boys $[p = (1/4)]$

or $1$ boy, $1$ girl $[p = (1/2)]$,

or $2$ girls $[p = (1/4)]$.

Therefore, when a boy answers on Monday, the chance that there are $2$ girls has been eliminated, and there are $2$ ways that the boy answers.

BB: probability $~= (1/4) \times 1 = (1/4)$.

BG: probability $~= p(1/2) \times p(1/2) = (1/4).$

Therefore, before you knock on the door on Tuesday, you conclude that it is equally likely that there are $2$ boys as that there is only $1$ boy.

Edit
Informally, I am using Bayes Theorem here.
However, I am focusing on the intermediate conclusion that BB and BG are now equally likely, where BG means exactly 1 boy.

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Therefore, on Tuesday, the chance of a boy answering is

BB: $(1/2) \times 1$.

BG: $(1/2) \times (1/2)$.

Therefore, overall chance of boy answering on Tuesday is $(3/4)$.

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Addendum
It is important to note that my knowledge of Probability theory has $50$ years of cobwebs, so I rely on intuition. Hence my use of an intermediate conclusion, while a more formal approach, used by someone with a clearer understanding of Probability theory, is often best.

On the other hand, the attack of such problems as this one are often facilitated by (what I refer to as) intermediate conclusions.

Answer 3

The only way a girl could have answered is for there to have been a BG pair (Pr =1/2) AND for the girl to have answered on the second day. (Pr=1/2)

Thus indicated Pr =1 - 1/4 =3/4

Answer 4

Almost the same logic with Monty Hall problem with 3 doors, (end up with 1/3 or 2/3). First time you visit a door and you know there is just 2 options 0.5 for a boy or 0.5 for a girl, After you open the door and notice a boy you actually fingure out that at least 1 of the 2 kids is a boy. So for the next day when you open the door you KNOW now that it's more chance for a boy. If we start with X1,X2 at the start now we know it's B,X2 (boy and the other kid could be a girl or a boy).

Now, If you really want to check if you understand try solving this: same problem but with 3 kids, first day you hit a boy what's the probability to find a boy the next day (if you can identify the boy that you have already saw). You can also thinking for a huge case such as N kids and what would happend if you come the X day knowing the results of the 1 to X-1 days.