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

Answer 1

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

Let the queue be $ABCDE$, starting from $A$.
So, girls can be positioned at $BD,BE,CD,CE$ and $DE$.
$5$ possibilities are there.

$\therefore$ required probability $=\frac{5\cdot 2!3!}{5!}=\frac{1}{2}$