Three boys and two girls stand in a queue. The probability, that the number of boys ahead is at least one more than the number of girls ahead of her, is `

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(2)/(3)`
D. `(3)/(4)`

Explanation

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) xx 2! xx 3! + .^(2)C_(1) xx 2! xx 3! = 36 + 24 = 60`.
Number of ways in which 3 boys and 2 girls can be seated = 5!
`therefore` Required probability = `(60)/(5!) = (1)/(2)`