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?

Let B represent a boy and G represent a girl. We have 3 boys and 2 girls. The total number of arrangements is given by the number of ways to arrange 5 children, which is 5!/(3!2!) = 10.

Let's list all possible arrangements and check the condition:

1. BBGBG: Girl 1: 2 boys, 0 girls ahead. Girl 2: 3 boys, 1 girl ahead. Condition satisfied.
2. BBGGB: Girl 1: 2 boys, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.
3. BGBBG: Girl 1: 1 boy, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.
4. BGBGB: Girl 1: 1 boy, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.
5. BGGBB: Girl 1: 0 boys, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.
6. GBBBG: Girl 1: 0 boys, 0 girls ahead. Girl 2: 3 boys, 1 girl ahead. Condition not satisfied.
7. GBBGB: Girl 1: 0 boys, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.
8. GBGBB: Girl 1: 0 boys, 0 girls ahead. Girl 2: 1 boy, 1 girl ahead. Condition not satisfied.
9. GGBBB: Girl 1: 0 boys, 0 girls ahead. Girl 2: 0 boys, 1 girl ahead. Condition not satisfied.
   10.BGBGB: Girl 1: 1 boy, 0 girls ahead. Girl 2: 2 boys, 1 girl ahead. Condition not satisfied.

Only arrangement 1 satisfies the condition. Therefore, there is only 1 arrangement that satisfies the given condition.

The probability is the number of favorable arrangements divided by the total number of arrangements.

Probability = 1/10

However, none of the options match this probability. There might be an error in the question or the options provided.