Question

$4$ Boys and $2$ girls occupy seats in a row at random. Then, the probability that the two girls occupy seat side by side is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is C

$\frac{1}{3}$

Explanation for the correct option:

Step 1: Find the number of arrangements for sitting of girls

Given: $4$ Boys and $2$ girls have to occupy random seats.

The number of arrangments of sitting of two girls among themself $=2!$

Step 2: Find the sample space and the favourable event

Total ways of sitting of $4$ Boys and $2$ girls $=5!2!$

So, $n\left(f\right)=5!2!$ will be favourable events.

Also, sample space, $n\left(s\right)=6!$

Step 3: Find the probability of two girls sitting together

The probability of two girls sitting together will be:

$\begin{array}{rcl}\frac{n\left(f\right)}{n\left(S\right)}& =& \frac{5!2!}{6!}\\ & =& \frac{1}{3}\end{array}$

Hence, option (C) is the correct answer.