Question

Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is

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

Answer 1

The correct option is D $\frac{1}{2}$

$n\left(S\right)=5!$
Case I: $G_1$
$2\times 4!=48$
Case II : $G_1$
$2\times 3!=12$
$n\left(E\right)=60$
$P\left(E\right)=\frac{60}{5!}=\frac{1}{2}$