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

$\text{Given: 3 boys and 2 girls stand in a queue}$

$\text{5 people in total}$

$\text{To find: } P\!\left(\#\text{boys ahead} \geq \#\text{girls ahead} + 1,\ \forall \text{ girls}\right)$

$(1)\ \frac{1}{2} \qquad (2)\ \frac{1}{3} \qquad (3)\ \frac{2}{3} \qquad (4)\ \frac{3}{4}$

$1.\ \text{total gender-patterns} = \binom{5}{2}$ $2.\ \text{score} = (\#\text{boys so far}) - (\#\text{girls so far})$ $3.\ \text{condition} \to \text{score path},\ \text{count favourable}$ $\text{Only the gender pattern matters (all B alike, all G alike)}$ $n(S) = \binom{5}{2} = 10$ $n(S) = 5!,\qquad n(F) = 5 \times 3! \times 2!$ $\dfrac{5\cdot 3!\cdot 2!}{5!} = \dfrac{5}{10} = \dfrac12$ $\text{Read left to right: boy} \to +1,\quad \text{girl} \to -1$ $\text{score starts at } 0$ $\text{For each girl: } (\#B - \#G)_{\text{ahead}} \geq 1$ $\Rightarrow \text{score before each girl} \geq 1 \Rightarrow \text{path never negative}$ $\text{1st person must be a boy (else } 0 \geq 1 \text{ fails for her)}$ $B\text{ fixed first: remaining } 4 \text{ slots hold } 2B,\,2G$ $\binom{4}{2} = 6 \text{ patterns}$ $B\,G\,G\,B\,B:\ \text{before 2nd girl, } \#B=1,\ \#G=1 \Rightarrow \text{score}=0 < 1\ \text{fail}$ $n(\text{favourable}) = 6 - 1 = 5$ $BGBGB,\quad BGBBG,\quad BBGGB,\quad BBGBG,\quad BBBGG$ $P = \dfrac{n(F)}{n(S)} = \dfrac{5}{10} = \dfrac12$ $\boxed{\text{Answer: (1)}\quad \dfrac12}$