Question

14. A jury pool consists of 26 people, 16 men, and 10 women. Compute the probability that a randomly selected jury of 12 people is all male.

Answer 1

Consider that the number of selections possible from 'n' distinct objects taken 'r' at a time is given by the formula,

$^nC_r=\frac{n!}{r!\cdot(n-r)!}$

As per the given problem, there are 26 people consisting 16 men and 10 women.

Then, the number of ways to select a jury of 12 people is calculated as,

$\begin{gathered} =^{26}C_{12} \\ =\frac{26!}{12!\cdot(26-12)!} \\ =\frac{26!}{12!\cdot14!} \end{gathered}$

Similarly, the number of ways to select a jury of 12 men is calculated as,

$\begin{gathered} =^{16}C_{12} \\ =\frac{16!}{12!\cdot(16-12)!} \\ =\frac{16!}{12!\cdot4!} \end{gathered}$

The probability of an event is given by,

$\text{Probability}=\frac{\text{ Number of favourable outcomes}}{\text{Number of total outcomes}}$

So the probability that the jury contains all 12 males is calculated as,

$\begin{gathered} P(\text{all males})=\frac{\text{ No. of ways of selecting jury containing all males}}{\text{ Total no. of ways of selecting the jury }} \\ P(\text{all males})=\frac{(\frac{16!}{12!\cdot4!})}{(\frac{26!}{12!\cdot14!})} \\ P(\text{all males})=\frac{16!}{12!\cdot4!}\cdot\frac{12!\cdot14!}{26!} \\ P(\text{all males})=\frac{16!}{4!}\cdot\frac{14!}{26!} \\ P(\text{all males})=\frac{7}{37145} \end{gathered}$

Thus, the required probability is obtained as,

$\frac{7}{37145}$