Probability of Relative Order of Boys and Girls in a Queue

To find the probability, we consider the relative order of the $3$ boys ($B$) and $2$ girls ($G$) in the queue of $5$ people.

The total number of ways to arrange $3$ boys and $2$ girls in a queue is given by the number of ways to choose the positions of the $2$ girls out of the $5$ available spots: $\text{Total outcomes} = \binom{5}{2} = \frac{5 \times 4}{2} = 10$

Let the positions in the queue from front to back be indexed as $1, 2, 3, 4, 5$.

Let the positions of the two girls be $p_1$ and $p_2$ such that $1 \le p_1 < p_2 \le 5$. The problem states that for every girl, the number of boys standing ahead of her must be at least one more than the number of girls standing ahead of her:

1. For the first girl (at position $p_1$):

   • Number of girls ahead of her $= 0$.
   • Therefore, the number of boys ahead of her must be at least $0 + 1 = 1$.
   • This implies $p_1 \ge 2$ (at least one boy must be ahead of her).

2. For the second girl (at position $p_2$):

   • Number of girls ahead of her $= 1$.
   • Therefore, the number of boys ahead of her must be at least $1 + 1 = 2$.
   • Since there is $1$ girl and at least $2$ boys ahead of her, her position must satisfy $p_2 \ge 1 + 2 + 1 = 4$.

Now, we list all possible pairs of positions $(p_1, p_2)$ that satisfy these conditions:

• If $p_1 = 2$:

  • $p_2 = 4$: The queue is $B \, G \, B \, G \, B$.
    • Ahead of $G_1$: $1$ boy, $0$ girls ($1 \ge 0 + 1$, valid).
    • Ahead of $G_2$: $2$ boys, $1$ girl ($2 \ge 1 + 1$, valid).
  • $p_2 = 5$: The queue is $B \, G \, B \, B \, G$.
    • Ahead of $G_1$: $1$ boy, $0$ girls ($1 \ge 0 + 1$, valid).
    • Ahead of $G_2$: $3$ boys, $1$ girl ($3 \ge 1 + 1$, valid).

• If $p_1 = 3$:

  • $p_2 = 4$: The queue is $B \, B \, G \, G \, B$.
    • Ahead of $G_1$: $2$ boys, $0$ girls ($2 \ge 0 + 1$, valid).
    • Ahead of $G_2$: $2$ boys, $1$ girl ($2 \ge 1 + 1$, valid).
  • $p_2 = 5$: The queue is $B \, B \, G \, B \, G$.
    • Ahead of $G_1$: $2$ boys, $0$ girls ($2 \ge 0 + 1$, valid).
    • Ahead of $G_2$: $3$ boys, $1$ girl ($3 \ge 1 + 1$, valid).

• If $p_1 = 4$:

  • $p_2 = 5$: The queue is $B \, B \, B \, G \, G$.
    • Ahead of $G_1$: $3$ boys, $0$ girls ($3 \ge 0 + 1$, valid).
    • Ahead of $G_2$: $3$ boys, $1$ girl ($3 \ge 1 + 1$, valid).

Thus, the number of favorable arrangements is $5$: $\{(2, 4), (2, 5), (3, 4), (3, 5), (4, 5)\}$

Therefore, the required probability is: $\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{10} = \frac{1}{2}$

Hence, the correct option is (A).