Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more than the number of girls ahead of her is

Question

Aicharya · Accepted Answer

Either a girl start the sequence or will be at second position and will not acquire the last position as well.
Required probability

= \frac{(^{3} C_{2} \times 2! \times 3! +^{2} C_{1} \times 2! \times 3!)}{5!} = \frac{1}{2}

Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is…

Solution