Question

After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only two women among the last 22 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 two or fewer women when 22 people are hired, assuming that there is no discrimination based on gender. (Report answer accurate to 8 decimal places). P(at most two) = .00021 Incorrect Because this is a serious claim, we will use a stricter cutoff value for unusual events. We will use 0.5% as the cutoff value (1 in 200 chance of happening by chance). With this in mind, does the resulting probability really support such a charge?

Answer 1

If there is no discrimination the probability of any one employee hired should be = 50% male and 50% female.

The binomial theorem uses the formula
P(exactly k successes from n events) = nCk* p(k)^k * ((1 -p(k))^(n-k)
where n is the number of events (22 people hired)
k is the number of successes (0,1 and 2 women respectively, in each of the three calcs)
p(k) is the probability any one employee hired is a woman = 1/2
1 -p(k) is the probability that any one employee hired is a man = 1/2
n - k = the number of men hired = 22 - (0,1,2) = 22,21,20 respectively in the three calcs

Therefore
P(excatly 0 women) = 22C0 * (1/2)^0 * (1/2)^22
As 22C0 and (1/2)^0 both equal 1, this becomes 1 * 1 * (1/2)^22
= (1/2)^22

P(exactly 1 woman) = 22C1 * (1/2)^1 * (1/2)^21
As 22C1 = 22! / (1!*21!) = 22
= 22* (1/2)^22

P(exactly 2 women) = 22C2 * (1/2)^2 *(1/2)^20
As 22C2 = 22! / (2!*20!!) = (22*21)/2 = 231
=231 *(1/2)^22

Therefore P(getting 2 or fewer women) = P(0) + P(1) + P(2)
= (1/2)^22 + 22*(1/2)^22 + 231*(1/2)^22
= 253*(1/2)^22

= 0.0000603199