1. Understanding the Problem and Sample Space

We have a queue of 3 boys (B) and 2 girls (G), for a total of 5 people. The total number of distinct arrangements (the size of our sample space, $n(S)$) is the number of permutations of these 5 people, where 3 are identical (B) and 2 are identical (G).

This can be calculated using the multinomial coefficient: $n(S) = \frac{5!}{3!2!} = \frac{120}{(6)(2)} = 10$ So, there are 10 possible unique arrangements for the queue.

2. Defining the Favorable Event

The condition for a favorable event is that for every girl in the queue, the number of boys ahead of her is at least one more than the number of girls ahead of her.

Let's formalize this. If a girl is at position $k$ in the queue (where $k=1, 2, 3, 4, 5$), let $N_B(k-1)$ be the number of boys in the first $k-1$ positions, and $N_G(k-1)$ be the number of girls in the first $k-1$ positions.

The condition is: For every position $k$ occupied by a girl, we must have: $N_B(k-1) \ge N_G(k-1) + 1$ This is equivalent to $N_B(k-1) > N_G(k-1)$.

3. Counting Favorable Outcomes

Instead of checking all 10 arrangements, it's often easier to count the unfavorable arrangements (the complement event, $E'$) and subtract from the total.

An arrangement is unfavorable if for at least one girl, the condition is not met, i.e., $N_B(k-1) < N_G(k-1) + 1$, or $N_B(k-1) \le N_G(k-1)$.

Let's identify the arrangements that are unfavorable:

Case 1: A girl is in the first position ($k=1$). If a girl is at the front of the queue, the number of people ahead of her is 0. So, $N_B(0) = 0$ and $N_G(0) = 0$. The condition becomes $0 \ge 0 + 1$, which is false. Therefore, any arrangement that starts with a girl is unfavorable.

How many such arrangements are there? If the first person is a girl, we need to arrange the remaining 3 boys and 1 girl in the other 4 positions. The number of ways to do this is: $\frac{4!}{3!1!} = 4$ These 4 arrangements are: GBBBG, GBGBB, GBBGB, GGBBB.

Case 2: The queue starts with a boy, but the condition fails later. Let's analyze prefixes of the queue. An arrangement is of the form B...

• If the second person is a girl (BG...), the girl is at $k=2$. Ahead of her is one boy ($N_B(1)=1, N_G(1)=0$). The condition is $1 \ge 0+1$, which is true. So this position is fine.
• Now consider the prefix BG. We have $N_B(2)=1$ and $N_G(2)=1$. If the third person ($k=3$) is a girl, the arrangement is BGG... Ahead of this second girl are one boy and one girl. The condition for her is $N_B(2) \ge N_G(2)+1$, which means $1 \ge 1+1$. This is false.
• So, any arrangement starting with BGG... is unfavorable. Since we have 3 boys and 2 girls in total, an arrangement starting with BGG must have the remaining two positions filled by boys. The only such arrangement is BGGBB.

Are there any other unfavorable arrangements starting with a boy? For the condition $N_B(k-1) \le N_G(k-1)$ to be met, the number of girls in any prefix must catch up to or exceed the number of boys.

• Start with B: $N_B > N_G$.
• Prefix BB: $N_B > N_G$.
• Prefix BG: $N_B = N_G$. This is a critical point. A girl cannot follow this prefix.
• Prefix BGB: $N_B > N_G$.
• Prefix BGBG: $N_B = N_G$. The next person is the 5th person. Since we have used 2 girls and 2 boys, the 5th person must be a boy. So the arrangement is BGBGB. This does not have a G following the critical prefix BGBG, so it's not unfavorable on this count.

So, the only unfavorable arrangements are the 4 that start with a girl and the 1 arrangement BGGBB.

Total number of unfavorable arrangements, $n(E')$, is $4 + 1 = 5$.

The number of favorable arrangements, $n(E)$, is: $n(E) = n(S) - n(E') = 10 - 5 = 5$

4. Calculating the Probability

The probability of the favorable event is: $P(E) = \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}} = \frac{n(E)}{n(S)} = \frac{5}{10} = \frac{1}{2}$

Conclusion

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 $\frac{1}{2}$. This corresponds to option A.