Family with two boys You call randomly a family with two kids, and ask if there is a kid called Tom. The answer is yes. Then what is the probably that the family has two boys.
So we want that $P(2 \text{ boys } | \text{ one of the children is called Tom })$. We can have:
\begin{array}{c}
Tom, B \\
B, Tom \\
Tom, G \\
G, Tom \\
\end{array}
So then since we know one of the kids is Tom then the probability must be $\frac{1}{2}$. Now if the question were to ask if we are given we know there is one boy and then we are asked to find the probability of two boys then we have:
\begin{array}{c}
B (known), B \\
B(known), G \\
G, G \\
\end{array}
So then this would be $P ( 2 \text{ boys } | \text{ one of them is a boy} ) = \frac{1}{3}$
Are these reasonings correct?
 A: Your answers are correct, but I'm not entirely convinced by your reasoning. Maybe it's my fault, but in any case I'll give another argument that someone might find useful.
Assume that the relative frequencies of boys and girls in the population are the same, namely $1/2$.
Let $t$ be the relative frequency of the name "Tom" in the population of boys.  Assume that no girls are named Tom.
Assume that for every couple, the sexes and names of that couple's children are drawn independently according to the distribution described above. Moreover, assume that $t$ is so small that $t^2 \approx 0$.  (This is one reasonable interpretation of the problem, I think.) Then by Bayes' theorem we have
$$P(2 \text{ boys} \mid \text{at least }1 \text{ Tom}) = \frac{P(\text{at least }1 \text{ Tom} \mid 2 \text{ boys})P(2 \text{ boys})}{P(\text{at least }1 \text{ Tom})} \approx \frac{2t \cdot 1/4}{t} = \frac{1}{2}.$$
As Alex points out in the comments below, if we calculate this exactly we get
\begin{align*}
P(2 \text{ boys} \mid \text{at least }1 \text{ Tom})
&= \frac{P(\text{at least }1 \text{ Tom} \mid 2 \text{ boys})P(2 \text{ boys})}{P(\text{at least }1 \text{ Tom})} \\
&= \frac{(1-(1-t)^2) \cdot 1/4}{1-(1-t/2)^2} = \frac{2-t}{4-t},
\end{align*}
and again the answer approaches $1/2$ as $t$ approaches zero.
For the second problem, the way I'd explain it is that before the phone is answered we have four equally likely possibilities: GG, GB, BG, and BB. After we get the information that it's not GG, then the remaining three are equally likely, and BB is one of these three, hence a probability of $1/3$.
Note that the answer to the first problem approaches the answer to the second problem as $t$ approaches $1$, because if all boys are named Tom, then the callers questions "...is named Tom" and "...is a boy" are equivalent.
A: Your reasoning is correct, if we assume that no parents are silly enough to name both of their children Tom.
