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

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=3C2×2!×3!+2C1×2!×3!=36+24=60.Number of ways in which 3 boys and 2 girls can be seated = 5!∴  Required probability =605!=12