Question

A group consists of 4 men, 3 women and 3 children. Three persons are selected from the
group at random. Find the probability that (i) at least 2 of them are children, and (ii) at the
most 2 of them are children.
11 119
[Ans.(i) (11/60) (ii) 119/120)​

Answer 1

Given :

No of men in the group = 4

No of women in the group = 3

No of children in the group = 3

From this group three persons are randomly selected .

To Find :

(i) probability that at least 2 of them are children .

(ii) probability that  at the  most 2 of them are children.

Solution :

Total no of persons in this group = 10

Total no of ways of selecting 3 persons from these 10 :

=$^{10}C_3$

=120

No of ways of selecting 3 persons with at least 2 children :

= with 2 children + with 3 children

=$^{3}C_2 \times ^{7}C_1 + ^{3}C_3$

= 22

So, the probability of selecting 3 persons with at least 2 children :

=$\frac{22}{120}$

=$\frac{11}{60}$

And no of ways of selecting 3 persons with at most 2 children :

= total ways - ways with 3 children

= 120 - $^{3}C_3$

= 120 - 1

= 119

So, the probability of selecting 3 persons with at most 2 children :

= $\frac{119}{120}$

Hence , the probability of selecting 3 persons with at least children and at most 2 children is $\frac{11}{60}$ and $\frac{119}{120}$ respectively .