Question

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

2.The probability that at least two female contestants will be among the three prize winners in an essay writing contest out of the six female and four male finalists.

3.The probability of getting a sum divisible by three if a pair of fair dice is rolled and the numbers that turn up are noted.

Answer 1

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 $\dbinom{6+6}{5}=792$ possible outcomes.

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

$$=\dfrac{1(6)}{792}=\dfrac{1}{132}$$ $$P(X=1)=\dfrac{\dbinom{6}{1}\dbinom{6}{5-1}}{\dbinom{12}{5}}$$

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

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

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

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

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

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

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

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

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

2. There are $\dbinom{6+4}{3}=120$ possible outcomes.

$$P(X\ge2)=P(X=2)+P(X=3)$$

$$=\dfrac{\dbinom{6}{2}\dbinom{4}{3-2}}{\dbinom{10}{3}}+\dfrac{\dbinom{6}{3}\dbinom{4}{3-3}}{\dbinom{10}{3}}$$ $$=\dfrac{15(4)}{120}+\dfrac{20(1)}{120}=\dfrac{2}{3}$$

3.

$$\begin{matrix} & 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 1+1 & 1+2 & 1+3 & 1+4 & 1+5 & 1+6 \\ 2 & 2+1 & 2+2 & 2+3 & 2+4 & 2+5 & 2+6\\ 3 & 3+1 & 3+2 & 3+3 & 3+4 & 3+5 & 3+6 \\ 4 & 4+1 & 4+2 & 4+3 & 4+4 & 4+5 & 4 +6 \\ 5 & 5+1 & 5+2 & 5+3 & 5+4 & 5+5 & 5+6 \\ 6 & 6+1 & 6+2 & 6+3 & 6+4 & 6+5 & 6+6 \end{matrix}$$

Total number of cases when two dice roles $=6×6=36$

The favorable outcomes are

$$1+2,1+5,2+1,2+4,3+3,3+6,$$

$$4+2,4+5,5+1,5+4,6+3,6+6$$

$$P(sum\ divisible\ by\ 3)=\dfrac{12}{36}=\dfrac{1}{3}$$