Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more that the number of girls ahead of her, is
- A.1/2
- B.1/3
- C.2/3
- D.3/4
Show answer and solution
Correct option: A
Solution
The condition requires that at every position a girl occupies, the number of boys before her exceeds the number of girls before her. This is equivalent to counting sequences of 3 B's and 2 G's where every prefix has more B's than G's — a classic ballot problem.
By the ballot problem result, the number of valid arrangements of m boys and n girls where boys always stay strictly ahead equals C(m+n, n) · (m−n+1)/(m+1). Here m=3, n=2: valid arrangements = C(5,2)·(2/4) = 10·(1/2) = 5.
Total arrangements = 5! / (3!2!) = 10, so probability = 5/10 = 1/2.
By the ballot problem result, the number of valid arrangements of m boys and n girls where boys always stay strictly ahead equals C(m+n, n) · (m−n+1)/(m+1). Here m=3, n=2: valid arrangements = C(5,2)·(2/4) = 10·(1/2) = 5.
Total arrangements = 5! / (3!2!) = 10, so probability = 5/10 = 1/2.
[JEE Main 2014]