Probability Ques 28
- 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) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $2 / 3$
(d) $3 / 4$
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Answer:
Correct Answer: 28.(a)
Solution:
Formula:
Total number of ways to arrange 3 boys and 2 girls is 5!.
According to the given condition, the 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 $=\frac{60}{120}=\frac{1}{2}$
BETA
