Question

An employer wishes to hire five people from a group of 24 applicants. 14 men and 10 women
all of whom are equally qualified to fill the position. If he selects the five at random, what is
the probability that (i) all five will be men, (ii) at least one will be a women

Answer 1

i) There are $C^{5}_{14}=\frac{14!}{5!(14-5!)}$ ways to choose 5 men out of 14.

Total number of ways of choosing 5 candidates out of 24 is $C^{5}_{24}=\frac{24!}{5!(24-5)!}$.

Thus, the probability that all five hired employees will be men equals $\frac{C^{5}_{14}}{C^{5}_{24}}=\frac{14!5!19!}{5!9!24!}=\frac{14\cdot13\cdot12\cdot11\cdot10}{24\cdot23\cdot22\cdot21\cdot20}=\frac{240240}{5100480}=0.047=4.7\%$.

ii) We can highlight 2 situations: 1) all of 5 employees are men; 2) NOT all of 5 employees are men, or in other words, there is at least 1 woman. Hence, the probability of the second situation $P_2=1-P_1$ , where $P_1$ is the probability that all of 5 are men. $P_1$ was found in i) and equals 0.047. Thus, the probability there will be at least 1 women among 5 new employees is $P_2=1-0.047=0.953=95.3\%$ .