Question

Three applicants are to be choosen at random out of 4 boys & 6 girls.what is the probability of selecting 1)all girls 2)all boys 3)at least one boy

Answer 1

There are $\binom {10}{3}=\frac{10!}{3!\cdot 7!}=\frac{10\cdot 9\cdot 8}{1\cdot 2\cdot 3}=120$ ways of choosing 3 applicants.

1) There are $\binom{6}{3}=\frac{6!}{3!\cdot 3!}=\frac{4\cdot 5\cdot 6}{1\cdot 2\cdot 3}=20$ ways of choosing 3 girls.

Then the probability of selecting all girls is $\frac{20}{120}=\frac{1}{6}$ .

2) There are $\binom{4}{3}=\frac{4!}{3!\cdot 1!}=4$ ways of choosing 3 boys.

Then the probability of selecting all boys is $\frac{4}{120}=\frac{1}{30}$ .

3) Events “all girls” and “at least one boy” are opposite. Therefore, the probability of selecting at least one boy is $1-\frac{1}{6}=\frac{5}{6}$ .

Answers: 1) $\frac{1}{6}$ ; 2) $\frac{1}{30}$ ; 3) $\frac{5}{6}$ .