Question

A committee of $4$ persons has to be chosen from $8$ boys and $6$ girls, consisting of at least one girl. Find the probability that the committee consists of more girls than boys.

Answer 1

Given $8$ boys and $6$ girls

Let $A$: selecting $4$ girls no boy

$B$: selecting $3$ girls and $1$ boy

$n\left(A\right)={{}^{6}C}_4.{{}^{8}C}_0=15\times 1=15$ ways

$n\left(B\right)={{}^{6}C}_3.{{}^{8}C}_1=20\times 8=160$ ways

$n\left(S\right)={{}^{14}C}_4=1001$ ways

$\therefore$ $P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{15}{1001}$ and $P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{160}{1001}$

Required probability$=P\left(A\right)+P\left(B\right)$

$\frac{15}{1001}+\frac{160}{1001}=\frac{175}{1001}$