Question 13 Marks
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?
Answer
View full question & answer→Let $b$ represent boy and $g$ represent girl. The sample space of experiment is
$
S=\{(b, b),(g, b),(b, g),(g, g)\}
$
Let E and F represent the following events respectively.
E 'Both the children are boy'
$F =$ 'At least one of the children is a boy'.
Then $\quad E =\{(b, b)\}$ and $F =\{(b, b),(g, b),(b, g)\}$
Now $\quad E \cap F =\{(b, b)\}$
So, $\quad P ( F )=\frac{3}{4}$ and $P ( E \cap F )=\frac{1}{4}$
$
\therefore \quad P(E \mid F)=\frac{P(E \cap F)}{P(F)}
$
Putting the values
$
P(E \mid F)=\frac{1 / 4}{3 / 4}
$
$P ( E \mid F )=\frac{1}{3}$
$
S=\{(b, b),(g, b),(b, g),(g, g)\}
$
Let E and F represent the following events respectively.
E 'Both the children are boy'
$F =$ 'At least one of the children is a boy'.
Then $\quad E =\{(b, b)\}$ and $F =\{(b, b),(g, b),(b, g)\}$
Now $\quad E \cap F =\{(b, b)\}$
So, $\quad P ( F )=\frac{3}{4}$ and $P ( E \cap F )=\frac{1}{4}$
$
\therefore \quad P(E \mid F)=\frac{P(E \cap F)}{P(F)}
$
Putting the values
$
P(E \mid F)=\frac{1 / 4}{3 / 4}
$
$P ( E \mid F )=\frac{1}{3}$