Question

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

Equation, expression, and inequalitie
ANSWERED
asked 2021-06-25

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: \(Pr(X=2)=\)\(\frac{\text{# of favorable outcomes}}{\text{ # of possible outcomes}}\) \(=\frac{30}{56} =\frac{15}{28}\) \(~ 0.5357 =53.57\%\)

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Best answer
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: \(\frac{\text{# of favorable outcomes}}{\text{ # of possible outcomes}}\) \(=30/56 =15/28 = 0.5357 =53.57%\)

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