First slide

Introduction to probability

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

Moderate

A

$\frac{1}{2}$
B

$\frac{1}{3}$
C

$\frac{2}{3}$
D

$\frac{3}{4}$

Solution

3 Boys and 2 Girls ...
(1) B (2) B (3) B (4)
Girl can't occupy 4th position.
Either girls can occupy 2 of 1, 2, 3 position or they can both be at position (1) or (2).
Hence, total number of ways in which girls can be seated
$=^{3} C_{2} \times 2! \times 3! +^{2} C_{1} \times 2! \times 3! = 36 + 24 = 60$ .
Number of ways in which 3 boys and 2 girls can be seated = 5!
$∴ Required probability = \frac{60}{5!} = \frac{1}{2}$

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