Question

A cricket club has $15$ members, among whom only $5$ can bowl. What is the probability of forming a team of $11$ to have at least $3$ bowlers?

Answer 1

Total number of ways of forming the team ${=}^{15}{C}_{11}{=}^{15}{C}_4$

Of these, number of ways of formation of the team

$\left(i\right)$with one bowler ${=}^5{C}_1{.}^{10}{C}_{10}=\frac{5!}{4!1!}\times \frac{10!}{10!}=5\times 1=5$

$\left(ii\right)$with two bowlers ${=}^5{C_2}^{10}{C}_9=\frac{5!}{3!2!}\times \frac{10!}{9!1!}=\frac{5\times 4\times 3!}{3!\times 2}\times \frac{10\times 9!}{9!}=5\times 2\times 10=100$

Probability that at least $3$ bowlers are in the team $=1-\frac{105}{{}^{15}{C}_4}=1-\frac{105}{\frac{15!}{11!4!}}=1-\frac{1}{13}=\frac{13-1}{13}=\frac{12}{13}$