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

Engineering

Mathematics

Conditional Probability

Question

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{3}{4}$

$\frac{2}{3}$

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possible cases

Case‑I: B G G B B

Case‑II: G G B B B

Case‑III: G B G B B

Case‑IV: B G B G B

Case‑V: G B B G B

Number of boys ahead of every girl is at least one more than the number of girls ahead of her, is $\frac{5 \times 3! \times 2!}{5!} = \frac{5 \times 6 \times 2}{120} = \frac{1}{2}$