Question

# There are 5 women and 3 men waiting on standby for a flight to New York.

Equation, expression, and inequalitie

There are 5 women and 3 men waiting on standby for a flight to New York. Suppose 3 of these 8 people are selected at random, and a random variable X is defined to be the number of women selected. Find $$Pr[X = 2]$$.

## Expert Answers (2)

2021-06-26

We need to select 3 of the 8 people. Since the order of the people is not important (as a different order results in the same people being selected), we need to use a combination.
Definition combination (order is not important);$$nCr=(n r)=\frac{n!}{r!(n-r)!}$$ with $$n!=n(n-1) \cdot \ldots \cdot 2 \cdot 1$$. # of possible outcomes=$$sC_3=\frac{8!}{3!(8-3)!}=\frac{8!}{3!5!}=56$$
We need to select 2 of the 5 women and 1 of the 3 men to obtain $$X=2$$. # of favorable outcomes=$$(5C_2 \cdot 3)C_1=(\frac{5!}{2!3!}) \cdot \frac{3!}{1!2!}=10 \cdot 3=30$$
The probability is the number of favorable outcomes divided by the number of possible outcomes: $$Pr(X=2)=$$$$\frac{\text{# of favorable outcomes}}{\text{ # of possible outcomes}}$$ $$=\frac{30}{56} =\frac{15}{28}$$ $$~ 0.5357 =53.57\%$$

2021-08-04

We need to select 3 of the 8 people. Since the order of the people is not important (as a different order results in the same people being selected), we need to use a combination.
Definition combination (order is not important);$$\displaystyle{n}{C}{r}={\left({n}{r}\right)}={n}\frac{!}{{{r}!{\left({n}-{r}\right)}!}}$$ with $$\displaystyle{n}\ne{n}{\left({n}-{1}\right)}\cdot\ldots{2}\cdot{1}$$. # of possible outcomes=$$\displaystyle{s}{C}{3}={}\frac{8!}{{{3}!{\left({8}-{3}\right)}!}}={}\frac{8!}{{3}}!{5}\ne{56}$$
We need to select 2 of the 5 women and 1 of the 3 men to obtain X=2. # of favorable outcomes=$$\displaystyle{\left({5}{C}_{2}\cdot{3}\right)}{C}_{1}={\left({}\frac{5!}{{2}}!{3}!\right)}\cdot{}\frac{3!}{{1}}!{2}\ne{10}\cdot{3}={30}$$
The probability is the number of favorable outcomes divided by the number of possible outcomes: $$\frac{\text{# of favorable outcomes}}{\text{ # of possible outcomes}}$$ $$=30/56 =15/28 = 0.5357 =53.57%$$