Question

• solve the question by (i) factorial method, (ii) using venn diagrm:
Q. An employer wishes to hire 3 people from a group of 15 applicants, 8 men and 7 women whome are equally qualified, to fill the position. If he selects the three at random, what is the probability that
(i) all three will be men
(ii) at least one will be a woman?

Answer 1

Answer on Question #58639 – Math – Statistics and Probability

Question

Solve the question by (i) factorial method, (ii) using Venn diagram.

An employer wishes to hire 3 people from a group of 15 applicants, 8 men and 7 women who are equally qualified, to fill the position. If he selects the three at random, what is the probability that

(i) all three will be men;

(ii) at least one will be a woman?

Solution

(i) $P = \left( \frac{\text{number of triples of men}}{\text{number of triples}} \right) = \frac{C_8^3}{C_{15}^3} = \frac{8!3!12!}{3!5!15!} = \frac{8 \cdot 7 \cdot 6}{15 \cdot 14 \cdot 13} = \frac{8}{65}$.

(ii) Let $S$ denote the set of all possible outcomes for the employer's selection. Let $A$ denote the subset of outcomes corresponding to the selection of three men and $B$ the subset corresponding to the selection of at least one woman.

$$\bar{B} = \{no\ women\} = A \Rightarrow A \cup B = S.$$ $$P(B) = 1 - P(A) = 1 - \frac{8}{65} = \frac{57}{65}.$$

Answer: (i) $\frac{8}{65}$; (ii) $\frac{57}{65}$.

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