I came across these two examples in Introduction to Probability - Blitzstein and I am confused what exactly is the difference between the two questions.
- Mr. Smith has two children. At least one of them is a boy. What is the probability that both the children are boys?
- A family has two children. You randomly run into one of the two, and learn that he is a boy. What is the conditional probability that both are boys?
Now, in the first question, we take
- A: Event that both children turn up to be boys
- B: Event that at least one of them is a boy.
We need to calculate $P(A|B)$ where $\Omega = \{BB,BG,GB,GG\}$ $$ P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(\{BB\})}{P(\{BB,BG,GB\})} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$ For the second question, we already have seen the first child which is B. The second child can either be B or G and the probability of that happening is $\frac{1}{2}$.
I fail to understand these different answers. In both the questions, the prior information we have is the same i.e. one of the children is B. Why are we getting different answers then?
Let $T$ be the event that there are two boys, $O$ be the event that we observed a boy and $A$ be the event that there is at least one boy.
Then we have $$P(T|O)=\frac{P(O|T)P(T)}{P(O)}$$
$$P(T|A)=\frac{P(A|T)P(T)}{P(A)}$$
It is easy to see that the numerators in those statements are the same, i.e. $(1)(\frac{1}{4})=\frac{1}{4}$. The difference lies in the denominators.
The probability that we observe a boy is $$P(O)=P(O|BB)P(BB) + P(O|BG)P(BG) + P(O|GB)P(GB) + P(O|GG)P(GG)$$
The probability that there is at least one boy is $$P(A)=P(A|BB)P(BB) + P(A|BG)P(BG) + P(A|GB)P(GB) + P(A|GG)P(GG)$$
Notice the events where there is one boy and one girl. The probability that we observe a boy in those cases is only $\frac{1}{2}$ but the probability that there is at least one boy in those cases is $1$.
Therefore, $P(A)>P(O)$ so $P(T|O)\gt P(T|A)$.