Question

After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only four women among the last 19 new employees. She also learns that the pool of applicants is very large, with an approximately equal number of qualified men as qualified women.
Help her address the charge of gender discrimination by finding the probability of getting four or fewer women when 19 people are hired, assuming that there is no discrimination based on gender.
(Report answer accurate to 8 decimal places).
P(at most four) =

Answer 1

Answer:

P(at most four) = 0.00960541

Step-by-step explanation:

For each employee hired, there are only two possible outcomes. Either it is a women, or it is not. The probability of an employee being a women is independent of any other employee, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

19 new employees.

This means that $n = 19$

Approximately equal number of qualified men as qualified women.

This means that $p = 0.5$

Probability of getting four or fewer women when 19 people are hired

This is:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{19,0}.(0.5)^{0}.(0.5)^{19} = 0.00000191$

$P(X = 1) = C_{19,1}.(0.5)^{1}.(0.5)^{18} = 0.00003624$

$P(X = 2) = C_{19,2}.(0.5)^{2}.(0.5)^{17} = 0.00032616$

$P(X = 3) = C_{19,3}.(0.5)^{3}.(0.5)^{16} = 0.00184822$

$P(X = 4) = C_{19,4}.(0.5)^{4}.(0.5)^{15} = 0.00739288$

Then

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.00000191 + 0.00003624 + 0.00032616 + 0.00184822 + 0.00739288 = 0.00960541$

So

P(at most four) = 0.00960541