Question

1. a 3 - member committee will be selected from a pool of 5 men and 7 women. create the probability distribution for the number of women on the committee. what is the expected number of women on the committee and what is the probability there are at least 2 women on the committee?

Explanation:

Step1: Define random variable

Let $X$ = number of women on the 3-member committee. $X$ can take values 0, 1, 2, 3.

Step2: Calculate total possible selections

Total ways to choose 3 people from 12:
$$\binom{12}{3} = \frac{12!}{3!(12-3)!} = 220$$

Step3: Find $P(X=0)$ (0 women, 3 men)

$$P(X=0) = \frac{\binom{7}{0}\binom{5}{3}}{\binom{12}{3}} = \frac{1 \times 10}{220} = \frac{10}{220} = \frac{1}{22}$$

Step4: Find $P(X=1)$ (1 woman, 2 men)

$$P(X=1) = \frac{\binom{7}{1}\binom{5}{2}}{\binom{12}{3}} = \frac{7 \times 10}{220} = \frac{70}{220} = \frac{7}{22}$$

Step5: Find $P(X=2)$ (2 women, 1 man)

$$P(X=2) = \frac{\binom{7}{2}\binom{5}{1}}{\binom{12}{3}} = \frac{21 \times 5}{220} = \frac{105}{220} = \frac{21}{44}$$

Step6: Find $P(X=3)$ (3 women, 0 men)

$$P(X=3) = \frac{\binom{7}{3}\binom{5}{0}}{\binom{12}{3}} = \frac{35 \times 1}{220} = \frac{35}{220} = \frac{7}{44}$$

Step7: Calculate expected value $E(X)$

$$E(X) = 0 \times \frac{1}{22} + 1 \times \frac{7}{22} + 2 \times \frac{21}{44} + 3 \times \frac{7}{44}$$
$$= 0 + \frac{7}{22} + \frac{42}{44} + \frac{21}{44} = \frac{14}{44} + \frac{42}{44} + \frac{21}{44} = \frac{77}{44} = \frac{7}{4} = 1.75$$

Step8: Calculate $P(X \geq 2)$

$$P(X \geq 2) = P(X=2) + P(X=3) = \frac{21}{44} + \frac{7}{44} = \frac{28}{44} = \frac{7}{11}$$

Answer:

Probability Distribution:

$X$ (Number of Women)	0	1	2	3

Expected Number of Women: $\frac{7}{4}$ or 1.75

Probability of at Least 2 Women: $\frac{7}{11}$