A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

$\frac{1}{3}$

The correct answer is Option (2) → $\frac{1}{3}$ ##

Let $g$ and $b$, respectively denote a girl and a boy.

So, sample space of the experiment is $S = \{(b, b), (g, b), (b, g), (g, g)\} [∴n(S) = 4]$

Let events $E = \text{Both the children are boys}$

$F = \text{At least one of the child is a boy}$

Then $E = \{(b, b)\} \quad [∴n(E) = 1]$

$F = \{(b, b), (g, b), (b, g)\} \quad [∴ n(F) = 3]$

Also, $E \cap F = \{(b, b)\} \quad [∴n(E \cap F) = 1]$

Thus, $P(F) = \frac{n(F)}{n(S)} = \frac{3}{4}$

and $P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{4}$

Therefore,

$P(E | F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$