Probability: find the probability of event B given that event A occurs My Problem,
Suppose a family has 2 children,
if one children is randomly selected and it is a girl then what is the probability of the second child to be a girl ?
Please help. Thanks
 A: Let's work by definition of conditional probabilities:
Let A = first child is girl, and B = second child is girl.
$$P(B \mid A) = \frac{P(A \cap B) }{P(A)}= \frac{0.5 \times 0.5}{0.5}=0.5$$
A: "if one children is randomly selected and it is a girl then what is the probability of the second child to be a girl"
This is one of the worst cases of ambiguity in a question that I've seen in a while.  "second child" usually means the one born after the first child, but that seems to conflict with the statement that one of the two is randomly chosen.  I have to construe "the second child" to mean "the other child".
$$
\begin{array}{ccl}
\text{first born} & \text{second born} \\
\hline
\text{boy} & \text{boy} & \longleftarrow\text{This case is ruled out.} \\
\text{boy} & \text{girl} \\
\text{girl} & \text{boy} \\
\text{girl} & \text{girl}
\end{array}
$$
If one of two randomly chosen is a girl, then in one of the three cases not ruled out, the other one is a girl.  So $1/3$ is the bottom line.
A: Let $A$ be the event that the you randomly selected a girl. 
Let $B$ be the event that both children are girls.
Let $C$ be the event that one child is a boy and one is a girl. 

$$P(A) = 1/2$$
$$P(B) = 1/4$$
$$P(C) = 1/2$$

What is $P(A \,\mid\,B)$? The probability that you randomly select a girl, given that both children are girls. This is clearly $1$. 

We are looking for $P(B \,\mid\,A)$. We can use Bayes' rule:
$$P(B\,\mid\,A) = \frac{P(A\,\mid\,B)\cdot P(B)}{P(A)}$$
$$P(B\,\mid\,A) = \frac{1 \cdot \frac{1}{4}}{\frac{1}{2}}$$
$$P(B\,\mid\,A) = \boxed{\frac{1}{2}}$$
