Question 1099641: i have solved b plz help w/ a and b
A company has five applicants for two positions: two women and three men.
Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender.
Let x equal to the number of women chosen to fill the two positions.
a) Write down the appropriate probability distribution with its parameters?
b) What are the mean and variance of this distribution?
The mean = 0.8
The standard deviation = 0.6
c) What is the probability that a company will hire exactly one woman?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have 2 positions to fill.
there are 5 candidates, 2 of which are women and 3 of which are men.
there is an equal probability that any of the candidates will be chosen, regardless of gender.
the probability that the positions will be filled with 2 women is:
c(2,2) / c(5,2) = 1 / 10 = 10%
the probability that the positions will be filled with 2 men is:
c(3,2) / c(5,2) = 3 / 10 = 30%
the probability that the position will be filled with 1 woman and 1 man is:
c(2,1) * c(3,1) / c(5,2) = (2 * 3) / 10 = 60%
the sum of all probabilities is equal to 100% as it should be.
c(n,x) = n! / (x! * (n-x)!)
this represents the number of ways you can select sets of x out of a set of n where order within each of those sets is not important.
for example:
c(5,2) = n! / (2! * 3!) = (5*4*3*2*1) / ((2*1)*(3*2*1)) = 120 / (2*6) = 120 / 12 = 10.
the probabilities can also be calculated as follows:
no replacement is assumed.
the probability that the first pick is a woman is 2/5.
the probability that the second pick is also a woman is 1/4.
the probability that both picks are women is 2/5 * 1/4 = 2/20 = 1/10 = 10%
the probability that the first pick is a woman is 2/5.
the probability that the second pick is a man is 3/4.
the probability that the first pick is a woman and the second pick is a man is 2/5 * 3/4 = 6/20 = 3/10 = 30%.
the probabillty that the first pick is a man is 3/5.
the probability that the second pick is a woman is 2/4.
the probability that the first pick is a man and the second pick is a women is 3/5 * 2/4 = 6/20 = 3/10 = 30%.
the probability that the first pick is a man is 3/5.
the probability that the second pick is also a man is 2/4.
the probability that the first pick is a man and the second pick is also a man is 3/5 * 2/4 = 6/20 = 3/10 = 30%.
the probability that both are women is 10%.
the probabibility that one is a man and one is a woman is 2 * 30% = 60%.
the probability that both are men is 30%.
you get the same probabilities at 10% and 60% and 30% for a total probability of 100%, same as the first analysis.
i'm not exactly sure what distribution they are talking about, and therefore not sure what the mean or the standard deviation of that distribution would be.
the probability that the company will hire exactly 1 woman for the 2 positions would be .6.
the only distribution that i can think of would be the probability distribution of number of women chosen for the position.
p(0 women) = .3
p(1 woman) = .6
p(2 women) = .1
if this is the distribution they are looking for, then:
the mean would be .33 rounded to 2 decimal places.
the standard deviation would be .21 rounded to 2 decimal places.
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