SOLUTION: At a community college, there are seven executive officers: four women and three men. Three are selected to attend the national accreditation convention. Find the probability tha
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Question 355171: At a community college, there are seven executive officers: four women and three men. Three are selected to attend the national accreditation convention. Find the probability that:
1) all three selected will be women
2) all three selected will be men
3) two men and one woman will be selected
4) one man and two women will be selected Found 2 solutions by ewatrrr, jrfrunner:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! Hi,
Note: The probability of x successes in n trials is:
.
P = nCx* * where p and q are the probabilities of success and failure respectively. In this case p(a women chosen) = 4/7 and q = 3/7
.
nCx =
.
1) P(all three selected will be women) =
.
2) P(all three selected will be men) =
.
3) P(two men and one woman will be selected) =
.
4) P(one man and two women will be selected) =
You can put this solution on YOUR website! This resembles a binomial experiment: two outcomes, choosing a Man or choosing a Woman. But it fails in that the probability of selecting a Man or a Woman is not constant so therefore, this is not a binomial experiment but a hypergeometric experiment
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given: N=7 officers, 4 are Women and 3 are Men, sample size n=3
==
1) all three selected will be women
Let X=number of women selected
There are 7C3 ways of selecting a group of 3 from the 7 choices
There are 4C3 ways of selecting 3 women from the 4 available
There are 3C0 ways of selecting 0 men from the 3 avialable
P(X=3)=(4C3*3C0)/7C3=4/35
==
2) all three selected will be men
Let X=number of women selected
There are 7C3 ways of selecting a group of 3 from the 7 choices
There are 4C0 ways of selecting 0 women from the 4 avialable
There are 3C3 ways of selecting 3 men from the 3 available
P(X=0)=(4C0*3C3)/7C3=1/35
==
3) two men and one woman will be selected
Let X=number of women selected
There are 7C3 ways of selecting a group of 3 from the 7 choices
There are 4C1 ways of selecting 1 woman from the 4 avialable
There are 3C2 ways of selecting 2 men from the 3 available
P(X=1)=(4C1*3C2)/7C3=12/35
==
4) one man and two women will be selected
Let X=number of women selected
There are 7C3 ways of selecting a group of 3 from the 7 choices
There are 4C2 ways of selecting 2 women from the 4 avialable
There are 3C1 ways of selecting 1 man from the 3 available
P(X=2)=(4C2*3C1)/7C3=6*3/35=18/35
