Probability that first born is a boy I had this question on an exam last week: there is a family, and they have $3$ children, you know that they have exactly $2$ boys and $1$ girl. What is the probability that the first born is a boy? 
Some people answered $1/2$ and others $2/3$, what do you think? We can't figure it out. 
 A: Possibilities are: bbg bgb gbb
Favorable: bbg bgb
A: The three different possibilities are BGB - BBG - GBB => Probability of a boy first is two thirds. 
Another way to look at it: call the children X,Y,Z, with Y and Z males, and X female. So what is the probability that X is first? It's 1/3, since they are all equally likely to come up first. So with probability 2/3 the first one is either Y or Z, i.e. a male.
A: We can use Bayes' Theorem for conditional probability problems:
$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$
In this case, the events A and B are:A: first born is boy
B: family have exactly two boys, one girl
Bayes' Theorem then states for this problem that:

The probability of the first born being a boy given that the family has two boys and a girl is:
the probability that the family has two boys and a girl given the first born is male, times the probability the first born is a boy, divided by the probability the family has two boys and a girl.

Each of the probabilities needed can be calculated:
$$
\begin{align}
P(B|A)&=P(\text{the family has two boys and a girl|the first born is male})=\frac12\\
P(A)&=P(\text{first born is a boy})=\frac12\\
P(B)&=P(\text{the family has two boys and a girl})=\frac38
\end{align}
$$
Putting these together gives:
$$P(A|B)=\frac{\frac12\frac12}{\frac38}=\frac23$$
