Question 1022633: A committee consists of 10 women and 7 men. Three members are chosen as officers. What is the probability that all three officers are women? What is the probability that one of the officers chosen is a woman and the other two are men? What is the probability that at lease one officer is a man?
Answer by mathmate(429)   (Show Source):
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Question:
A committee consists of 10 women and 7 men. Three members are chosen as officers.
(a) What is the probability that all three officers are women?
(b)What is the probability that one of the officers chosen is a woman and the other two are men?
(c) What is the probability that at lease one officer is a man?
Solution:
Given 10 women and 7 men, from which 3 are chosen as officers.
Use C(n,r)=n!/(r!(n-r)!) to represent number of combinations of taking r objects out of n.
(a)All 3 are women
Ways to choose 3 women out of 7 = C(7,3)
Ways to choose from 17 members = C(17,3)
Probability of choosing 3 women out of 17
= C(10,3)/C(17,3)
= 120/680
= 3/17
=0.17647 (approx.)
(b) One woman and two men
Ways to choose 1 woman out of 10 = C(10,1) = 10
Ways to choose 2 men out of 7 = C(7,2) = 21
Ways to choose 1 woman and 2 men = 10*21=210
Probability of choosing 1 woman and 2 men = 210/680 = 21/68 = 0.3088 approx.
(c) At least one officer is a man.
This is the complement of case (a), therefore, the
Probability = 1-3/17 = 14/17 = 0.82353 (approx.)
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