Question

2n boys are divided into two sub-groups containing n boys each. The probability that the two tallest boys are in different groups is

(a) \(\frac n{2n-1}\)

(b) \(\frac {n-1}{2n-1}\)

(c) \(\frac {2n-1}{4n^2}\)

(d) None of these

Answer 1

Correct option is (a) \(\frac n{2n-1}\) 

Total number of ways of dividing 2n boys in two equal groups is ²ⁿC_n . ⁿC_n.

If we leave the two tallest boys, 2n–2 boys can be divided into two equal groups in ^2n–2C_n–1.

^n–1C_n–1 ways and the two tallest boys can be arranged in different groups in 2! ways. 

\(\therefore\) The required probability is