Answer to Question #305475 in Statistics and Probability for BELLAFAYRE

Question

The number of females employees selected if five employees are selected by lottery from a group of six male and six female employees.

Answer 1

Let $X=$ the number of females employees selected. The possible values of the random variable $X$ are $0, 1, 2,3,4,5.$

There are $\dbinom{6+6}{5}=792$

$$P(X=0)=\dfrac{\dbinom{6}{0}\dbinom{6}{5-0}}{\dbinom{12}{5}}$$

$$=\dfrac{1(6)}{792}=\dfrac{1}{132}$$

$$P(X=1)=\dfrac{\dbinom{6}{1}\dbinom{6}{5-1}}{\dbinom{12}{5}}$$

$$=\dfrac{6(15)}{792}=\dfrac{15}{132}$$

$$P(X=2)=\dfrac{\dbinom{6}{2}\dbinom{6}{5-2}}{\dbinom{12}{5}}$$

$$=\dfrac{15(20)}{792}=\dfrac{50}{132}$$

$$P(X=3)=\dfrac{\dbinom{6}{3}\dbinom{6}{5-3}}{\dbinom{12}{5}}$$

$$=\dfrac{20(15)}{792}=\dfrac{50}{132}$$

$$P(X=4)=\dfrac{\dbinom{6}{4}\dbinom{6}{5-4}}{\dbinom{12}{5}}$$

$$=\dfrac{15(6)}{792}=\dfrac{15}{132}$$

$$P(X=5)=\dfrac{\dbinom{6}{5}\dbinom{6}{5-5}}{\dbinom{12}{5}}$$

$$=\dfrac{6(1)}{792}=\dfrac{1}{132}$$