Quantitative Reasoning

Probability

Medium

Four people are chosen at random from a group of 3 men, 2 women and 4 children. The probability that exactly 2 of them are children is:

$\frac{10}{21}$

$\frac{1}{9}$

$\frac{1}{12}$

$\frac{1}{5}$

✅ Correct Option: 1

The group contains:

Men = 3

Women = 2

Children = 4

Total people = 3 + 2 + 4 = 9

The total number of ways to choose 4 people from 9:

$C(9,4) = \frac{9!}{4! \times 5!}$

$= \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

$= \frac{3024}{24}$

$= 126$

For exactly 2 children, the other 2 must be non-children (men or women).

Non-children = 3 + 2 = 5 people

Choosing 2 children from 4:

$C(4,2) = \frac{4 \times 3}{2 \times 1} = 6$

Choosing 2 non-children from 5:

$C(5,2) = \frac{5 \times 4}{2 \times 1} = 10$

Total favorable outcomes = $6 \times 10 = 60$

The probability is:

$P = \frac{60}{126}$

$= \frac{10}{21}$

Therefore, the probability that exactly 2 of them are children is $\frac{10}{21}$ .

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Solution

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