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 t
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Question 1167853: 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. Both candidate selected are women?
ii. At least one candidate selected is a woman?
iii. Second candidate is a woman.
iv. First candidate is a woman given that the second candidate 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. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 5 W
3 M
both women is (5/8)(4/7)=20/56 and can be reduced or put into decimal form.
none is a woman is (3/8)(2/7)=6/56. So at least 1 is a woman is the complement or 50/56.
second candidate is a woman. The first is 5/8 and the second is 4/7 for 20/56 as above
the first is a man and the second is a woman is 3/8 and 5/7 or 15/56
The sum is 35/56.
Given that the second candidate is a woman (35/56) the probability the first is...
we know that the first and second was 20/56, so the answer is 20/35 or 4/7.
0 women is 6/56
1 woman is 30/56
2 women is 20/56
The expected value is 30/56 + 40/56 or 70/56 or 1.25 women. That is the mean
the sd is sqrt of variance. The variance is 1.25^2(6/56)+.25^2(30/56)+0.75^2(20/56)
0.1674+0.3348+0.2009=0.7031.
the sd is sqrt(0.7031) or 0.8385
