Question

A group contains $10$ men and $4$ women. Three members committee is formed which must contain at least on woman. Find the probability that the committee so formed has more women and men.

Answer 1

Total no. of ways forming 3 member committee having

at least 1 woman = ${}^4{C}_1{\times }^{10}{C}_2{+}^4{C}_2{\times }^{10}{C}_1{+}^4{C}_3{\times }^{10}{C}_0$

$=45\times 4+6\times 10+4\times 1=180+60+4$

$=135+30+1=166=224$

No. of ways in which committee consists of more women than man

=[2 woman + 1 man]+ [3 women]

${=}^4{C}_2{\times }^{10}{C}_1{+}^4{C}_3{\times }^{10}{C}_0$

$=6\times 10+4\times 1=64$

Required Probability = $\frac{64}{2\times 4}=2/7$