Answer to Question #183300 in Statistics and Probability for Muhammad Haris

Question

Out of 25 employees of a company, 5 are engineers. Three employees are selected at random for granting leave. What is the probability that (i) all the three are engineers?(ii) none of them is an  engineer? (iii) at least one of them is an engineer

Answer 1

The total number of ways to select three employees out of 25 employees is

$$\dbinom{25}{3}=\dfrac{25!}{3!(25-3)!}=2300$$

(i)

$$P(3\ engineers)=\dfrac{\dbinom{5}{3}\dbinom{25-5}{3-3}}{\dbinom{25}{3}}=\dfrac{10\cdot1}{2300}$$ $$=\dfrac{1}{230}$$

(ii)

$$P(0\ engineers)=\dfrac{\dbinom{5}{0}\dbinom{25-5}{3-0}}{\dbinom{25}{3}}=\dfrac{1\cdot1140}{2300}$$ $$=\dfrac{57}{115}$$

(iii)

$$P(at\ least \ 1\ engineer)=1-P(0\ engineers)$$ $$=1-\dfrac{57}{115}=\dfrac{58}{115}$$