Question

From a group of 8 boys and 5 girls, a committee of 5 is to be formed. Find the probability that the committee contains
a) 3 boys and 2 girls
b) at least 3 boys.

Answer 1

Solution:

To form a team of 5 members

1) Probability of choosing 3 boys and 2 girls:

Favourable outcomes to choose 2 girls from 5
$\rm5C_{2}$
Favourable outcome to choose 3 boys from 8

$\rm8C_{3}$
Total outcomes to chose 5 members from 13

$\rm13C_{5}$
Let the event of choosing 3 boys and 2 girls is Z,so probability

$p(Z) = \frac{\rm5C_{2} \times \rm8C_{3}}{\rm13C_{5}} \\ \\ = \frac{ \frac{5!}{3! \: 2!} \times \frac{8!}{5 !\: 3!} }{ \frac{13!}{8! \: 5!} } \\ \\ = \frac{10 \times 56 }{1287} \\ \\ = \frac{560}{1287} \\ \\ p(Z)= 0.435$

2) For at least three boys:

Team can be choosing as

3 boys, 2 girls: p(Z)= 0.435

4 boys,1 girl

5 boys

for 4 boys and 1 girl,let this event is X
$p(X) = \frac{\rm5C_{1} \times \rm8C_{4}}{\rm13C_{5}}\\ \\ = \frac{ \frac{5!}{4! \: 1!} \times \frac{8!}{4!\: 4!} }{ \frac{13!}{8! \: 5!}} \\ \\ = \frac{5 \times 70 }{1287} \\ \\ = \frac{350}{1287} \\ \\ p(X)= 0.2719$

Let the event of choosing 5 boys for team is Y

So
$p(Y) = \frac{ \rm8C_{5}}{\rm13C_{5}} \\ \\ = \frac{\frac{8!}{5! \: 3!} }{ \frac{13!}{8! \: 5!} } \\ \\ = \frac{ 56 }{1287} \\ \\ = \frac{56}{1287} \\ \\ p(Y)= 0.0435$
So,probability of choosing at least 3 boys be B

$p(B)=p(Z)+p(X)+p(Y)\\\\p(B)= 0.435 + 0.2719 + 0.0435 \\ \\ p(B)= 0.75$
Hope it helps you.