Probability 1 Question 8

8. Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more that the number of girls ahead of her, is (2014 Adv)

(a) $1 / 2$

(b) $1 / 3$

(c) $2 / 3$

(d) $3 / 4$

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Answer:

Correct Answer: 8. (a)

Solution:

  1. Total number of ways to arrange 3 boys and 2 girls are 5 !.

According to given condition, following cases may arise.

$B$ $G$ $G$ $B$ $B$
$G$ $G$ $B$ $B$ $B$
$G$ $B$ $G$ $B$ $B$
$G$ $B$ $B$ $G$ $B$
$B$ $G$ $B$ $G$ $B$

So, number of favourable ways $=5 \times 3 ! \times 2 !=60$

$\therefore \quad$ Required probability $=\frac{60}{120}=\frac{1}{2}$



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