Question

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?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{1}{2}$

The total number of ways to arrange $3$ boys and $2$ girls is $5!=5\times 4\times 3\times 2\times 1=120$.
According to given condition, the following cases may arise.
Order of standing: Left to right
$\begin{array}{ccccc}B& G& B& G& B\\ B& B& G& B& G\\ B& B& B& G& G\\ B& G& B& B& G\\ B& B& G& G& B\end{array}$
Total cases $=5$ ; Arrangement of 3 boys and 2 girls in each case $=3!\times 2!=3\times 2\times 2=12$
So, number of favourable ways $=5\times 3!\times 2!=5\times 12=60$
$\therefore$ Required probability $=\frac{60}{120}=\frac{1}{2}$