If four persons are chosen at random from a group of 3 men,... | MATH - USE OF PERMUTATIONS AND COMBINATIONS IN PROBABILITY | SolveeIt | Solveeit

Today, August 18

Question

If four persons are chosen at random from a group of 3 men, 2 women and 4 children. Then the probability that exactly two of them are children, is

A

$\frac { 10 } { 21 }$

B

$\frac { 8 } { 63 }$

C

$\frac { 5 } { 21 }$

D

$\frac { 9 } { 21 }$

Answer

$\frac { 10 } { 21 }$

Explanation

Solution

Total number of ways $= { } ^ { 9 } C _ { 4 }$ 2 children are chosen in ${ } ^ { 4 } C _ { 2 }$ ways and other 2 persons are chosen in ${ } ^ { 5 } C _ { 2 }$ ways.

Hence required probability =