SOLUTION: A company is to hire two new employees. They have prepared a final list of eight candidates, all of whom are equally qualified. Of these eight candidates, five are women. Suppose

Question 1167849: A company is to hire two new employees. They have prepared a final list of eight candidates, all
of whom are equally qualified. Of these eight candidates, five are women. Suppose the
company decides to select two persons randomly from these eight candidates.
a. What is the probability that:
i. At least one candidate selected is a woman?

b. Let X denote the number of women in this sample.
i. Write the probability distribution of X.
ii. Find the standard deviation of X.
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

There are four elements in the sample space:

2 men 3/8 X 2/7 6/56 1 man/1 woman 3/8 X 5/7 15/56 1 woman/1 man 5/8 X 3/7 15/56 2 women 5/8 X 4/7 20/56

The answer is either the sum of the last three elements or one minus the first element.

John

My calculator said it, I believe it, that settles it

I > Ø

Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!

A company is to hire two new employees. They have prepared a final list of
eight candidates, all of whom are equally qualified. Of these eight candidates,
five are women.
Suppose the company decides to select two persons randomly from these eight
candidates.
a. What is the probability that:
i. At least one candidate selected is a woman?
Since 5 are women, 3 are men. We get the probability of the complement event and subtract from 1:
1 - (the probability that both are men) = 1 - C(3,2)/C(8,2) = 1 - 3/28 = 28/28 - 3/28 = 25/28
b. Let X denote the number of women in this sample.
i. Write the probability distribution of X.
X P(X) ---------------------------------------------- 0 [C(5,0)C(3,2)]/C(8,2) = (1)(3)/28 = 3/28 1 [C(5,1)C(3,1)]/C(8,2) = (5)(3)/28 = 15/28 2 [C(5,2)C(3,0)]/C(8,2) = (10)(1)/28 = 10/28 ii. Find the standard deviation of X. x p x∙p x² x²∙p ---------------------------------------------- 0 3/28 0 0 0 1 15/28 15/28 1 15/28 2 10/28 20/28 4 40/28 ------------------------------- E(x)=35/28 E(x²)=55/28 Edwin

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