# Conditional probability with Bayes' Rule

On a practice exam from statistics I encountered a very difficult exercise I couldn't manage to solve:

In the tent next to you there is a family with two children. Early in the morning you see a boy coming out of the tent. What is the probability that the other child is a girl?

Use Bayes' Rule

My approach to the solution was the following:

We assume $P(GIRL)$ = 0.5 and similarly $P(BOY)$ = 0.5.

We have to compute the following conditional probability: $P($One child is a girl| One child is a boy).

By applying Bayes' rule we should be able to compute this probability.

Bayes Rule: $P(A|B)$ $=$ $\frac{P(B|A)*P(A)}{P(B|A)*P(A) + P(B|A^c)*P(A^c)}$

-
There is not enough information to solve the problem as stated. hejseb's answer is correct if we consider the information "you see a boy coming out of the tent" to be equivalent to the information "there is at least one boy in the family" (or at least equivalent conditional on your having observed the tent.) But perhaps it is more likely to see a boy coming out of the tent if there are two boys than it is to see a boy coming out of the tent if there is only one boy. – Trevor Wilson Oct 24 '13 at 20:14

You can obtain two answers to this, actually. The problem is called the Sisters' Paradox. See this excellent explanation.

The most common solution, I would say, goes as follows. Let $G$ denote a girl, and $B$ a boy such that $P(BG)$ means probability of a girl and a boy. $P(GG)=P(BB)=1/4$, $P(BG)=1/2$. Conditioning on a boy($P(B)$):

$$P(BG|B)=\frac{P(BG)}{P(B)}=\frac{P(BG)}{P(BB)+P(BG)}=\frac{1/2}{1/2+1/4}=\frac{2}{3}$$

Note that in the reference this way of solving the question yields $1/3$, but that's because in that case it's $P(GG|G)$ (or, equivalently, $P(BB|B)$) rather than one of each. But, $P(BB|B)=1-P(BG|B)=1-2/3=1/3$, so the answers are in spirit the same - just different formulations.

-

The answer is simply 1/2. If you are required to use conditional probability, it goes like this: $$P(G_2 | B_1) = \frac{P(G_2 \cap B_1)}{P(B_1)} = \frac{P(G_2) P(B_1)}{P(B_1)} = \frac{(1/2)(1/2)}{1/2} = 1/2$$

Don't confuse this question with a very similar one where all we know at the start is that at least one child is a boy. If you have seen one child, and that one is a boy, it has no effect on the probability that the remaining child is a girl; they are independent events.

 Despite being down-voted on this one, I am sticking to my guns. Here is another way to look at it. Initially, the possibilities are {BB, BG, GB, GG}, where the first letter denotes the first child seen and all possibilities are equally likely, by assumption. After the first child is seen the remaining possibilities are {BB, BG}, each of which is equally likely. The event that the second child is a girl corresponds to BG, with probability 1/2.

I repeat, don't confuse this with the problem where all we know is that at least one child is a boy. In that case, after elimination the possibilities are {BB, BG, GB]; but that's different.[/edit]

-
This answer is wrong. You oversimplificated the calculation of some of your probabilities. Please refer to hejseb's answer; the answer to this question should be $\frac{2}{3}$. – dreamer Jul 6 '13 at 10:37
See my expanded answer above. – awkward Jul 6 '13 at 15:05
Your answer does not take into account the fact that $GB$ (as in the girl is the oldest) would also be a possible outcome. The order in which the children were born and that in which they leave the tent should be clearly distinguished between. – dreamer Jul 6 '13 at 15:12
Where do you see "oldest" in the problem statement? We saw one child, and that child was a boy. I believe there is a good discussion here: mathforum.org/dr.math/faq/faq.boygirl.choose.html In particular, the section titled "Choosing the child first". – awkward Jul 6 '13 at 15:31
Another argument, what makes you think that $G_2 \text{ and } B_1$ are independent? Those are assumptions you cannot just make. If we could make such assumptions the whole exercise would have been pointless in the first place. – dreamer Jul 6 '13 at 15:44

Let $S$ refer to the child seen. Let $BG$ refer to the event that the family has one boy and one girl (i.e. I don't mean to distinguish $BG$ and $GB$). Then a minor variant of Bayes' rule gives

$$P(BG | S=B) = \frac{P(S=B | BG) P(BG)}{ P(S=B | BG) P(BG) + P(S=B | BB) P(BB) + P(S=B | GG) P(GG)} = \frac{P(S=B | BG) \cdot \frac{1}{2}}{P(S=B | BG) \cdot\frac{1}{2} + 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4}} = \frac{2 P(S=B | BG)}{2 P(S=B | BG) + 1}$$

so to answer the question, you need to decide upon the probability that you would see the boy if it happened to be boy-girl.

If you decide $P(S=B|BG) = 1/2$, then $P(BG|S=B) = 1/2$.

What you tried to compute is incorrect;

$$P(\text{one is a girl} | \text{one is a boy}) = \frac{P(\text{one is a girl and one is a boy})}{P(\text{one is a boy})} = \frac{P(BG)}{P(BG) + P(BB)} = \frac{1/2}{1/2 + 1/4} = \frac{2}{3}$$

The problem is that this approach discards information: specifically that one child was seen, and it was that child that was a boy.

More explicitly, this calculation also includes the possibility that the family has one boy and one girl, and you saw a girl come out of the tent in the morning.

-