Question

A committee of 5 students is selected at random from a group consisting 10 boys and 5 girls. Given that there is at least one girl in the committee, calculate the probability that there are exactly 2 girls in the committee.

• A. $\frac{400}{917}$
• B. $\frac{400}{1001}$
• C. $\frac{143}{143}$
• D. $\frac{400}{991}$

Answer 1

The correct option is A $\frac{400}{917}$

$\mathrm{Let}E\mathrm{denotes}\mathrm{the}\mathrm{event}‘\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{girls}\mathrm{in}\mathrm{the}\mathrm{committee}’.\mathrm{and}F\mathrm{denotes}‘\mathrm{there}\mathrm{is}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{in}\mathrm{the}\mathrm{committee}’.\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{calculate}P\left(E|F\right).$

$A\mathrm{committee}\mathrm{of}5\mathrm{students}\mathrm{from}a\mathrm{group}\mathrm{consisting}\mathrm{of}10\mathrm{boys}\mathrm{and}5\mathrm{girls}\mathrm{can}\mathrm{be}\mathrm{formed}\mathrm{in}{15}_{C_5}\mathrm{ways}.\therefore \mathrm{The}\mathrm{number}\mathrm{of}\mathrm{elements}\mathrm{in}\mathrm{sample}\mathrm{space}={15}_{C_5}\mathrm{Since}F\mathrm{denotes}\mathrm{that}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{girl}\mathrm{is}\mathrm{chosen},F^C\mathrm{denote}\mathrm{that}\mathrm{no}\mathrm{girl}\mathrm{is}\mathrm{chosen},i.e.,\mathrm{five}\mathrm{boys}\mathrm{are}\mathrm{chosen}.\mathrm{Then}P\left(F^C\right)=\frac{{10}_{C_5}}{{15}_{C_5}}=\frac{12}{143}\therefore P\left(F\right)=1-\frac{12}{143}=\frac{131}{143}E\cap F\mathrm{will}\mathrm{consists}\mathrm{of}\mathrm{those}\mathrm{elements}\mathrm{where}\mathrm{there}\mathrm{are}2\mathrm{boys}\mathrm{and}3\mathrm{girls}.\therefore n\left(E\cap F\right)=5_{C_2}\times {10}_{C_3}=1200P\left(E\cap F\right)=\frac{5_{C_2}\times {10}_{C_3}}{{15}_{C_5}}=\frac{400}{1001}$
$\mathrm{Then},P\left(E|F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}=\frac{\frac{400}{1001}}{\frac{131}{143}}=\frac{400}{917}$