Question

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 1

Hint: For solving this question we will first find the total number of possible cases and then find the number of favourable cases and then, we will calculate the probability. Moreover, we will also use the concept of conditional probability here to find the correct answer.

Complete step-by-step answer:

Given:
A family has only 2 children.

Let, $A$ denote the event such that both children are boys and $B$ denote the event such that at least one of them is a boy.

In this question also we have to find the probability for the case such that both children are boys given that at least one of them is a boy. So, we are asked to find the value of $P\left( \dfrac{A}{B} \right)$ , where $P\left( \dfrac{A}{B} \right)$ represents the probability of an event $A$ when it is given that event $B$ has already happened.

Now, we will find the sample space for the given conditions. Every child has only two choices either to be a girl or a boy. Let, $S$ represents the sample space and $b$ stand for the boy and $g$ stand for the girl. Then, $S=\left\{ \left( g,g \right),\left( b,b \right),\left( b,g \right),\left( g,b \right) \right\}$

From $S$ we can write, $A=\left\{ \left( b,b \right) \right\}$ and $B=\left\{ \left( b,b \right),\left( b,g \right),\left( g,b \right) \right\}$ . Then,
$A\bigcap B=\left\{ \left( b,b \right) \right\}$

Now, if the sample space of $A$ and $B$ is the same. Then,
$P\left( \dfrac{A}{B} \right)=\dfrac{n\left( A\bigcap B \right)}{n\left( B \right)}..............\left( 1 \right)$

Where, the number of elements in $A\bigcap B=n\left( A\bigcap B \right)=1$ , and the number of elements in $B=n\left( B \right)=3$ .

Now, substituting the values of $n\left( A\bigcap B \right)$ and $n\left( B \right)$ in $\left( 1 \right)$ , we get:
$\begin{align}
  & P\left( \dfrac{A}{B} \right)=\dfrac{n\left( A\bigcap B \right)}{n\left( B \right)}=\dfrac{1}{3} \\
 & \Rightarrow P\left( \dfrac{A}{B} \right)=\dfrac{1}{3} \\
\end{align}$

Thus, $\dfrac{1}{3}$ is the probability for the case such that both the children are boys given that at least one of them is a boy.

Note: For solving any problem related to probability every student must try to make the cases according to the given constraints in the problem and then apply the formula accurately to get the correct answer. Conditional probability concepts should not be applied in the wrong way otherwise the answer will be wrong.