JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Probability

Total arrangements: $n(S)=5!$.

For the condition to be satisfied, each girl must have at least one more boy ahead of her than girls ahead of her. This means:

• The first girl must have at least 1 boy ahead
• The second girl must have at least 2 boys ahead (since she has 1 girl ahead)

The favorable arrangements can be counted by fixing the positions of boys and girls such that the condition is met.

Number of favorable arrangements: $n(E)=3!\times 2!\times 5=60$.

Probability: $P(E)=\frac{60}{120}=\frac{1}{2}$.