Question

2n boys are randomly divided into two subgroups containint n boys each. The probability that eh two tallest boys are in different groups is n/(2n-1) b. (n-1)/(2n-1) c. (n-1)/4n2 d. none of these A. n 2n-1 B. n-1 2n-1 C. 2n-1 4n2 D. none of these Select the correct answer from above options

Answer 1

Correct Answer - A The total number of ways of dividing 2n boys equally into a first and then a second subgroup is . 2 n C n × . n C n = . 2 n C n = 2 n ! n ! n ! . There are 2 ways in which the two tallest boys lie in different groups and corresponding to each way the remaining (2n-2) boys can be divided into two groups is . 2 n − 2 C n − 1 . . ∴ ∴ Favourable number of ways = 2 × . 2 n − 2 C n − 1 = Required probability = 2 × . 2 n − 2 C n − 1 ( 2 n ) ! n ! n ! = n 2 n − 1 =