Question

Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is
1. \(\frac{2}{9}\)
2. \(\frac{4}{5}\)
3. \(\frac{7}{12}\)
4. \(\frac{10}{21}\)

Answer 1

Correct Answer - Option 4 : \(\frac{10}{21}\)

Explanation:

Let ‘S’ be the sample space and ‘E’ be the event of choosing four persons such that 2 of them are children. Then,

⇒ n(S) = number of ways of choosing 4 persons out of 9

As we know the number of ways of selecting r things out of n is equal to nCr.

And   nCr  =  n!/(r! × (n − r)!)

 = 9C4 =(9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 126

n(E) = Number of ways of choosing 2 children out of 4 and 2 persons out of (3+2) persons

= (4C2 × 5C2) = [(4 × 3)/(2 × 1)] × [(5 × 4)/(2 × 1)] = 60

∴ P(E) = n(E)/n(S) = 60/126 = 10/21