Question 1120438: Three people are selected at random from six females and eight males. Find the probability of the following.
(a) At least one is a male.
(b) At most two are male.
Answer by greenestamps(13200)   (Show Source):
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There are only four possibilities: 3 male, 2 male and 1 female, 1 male and 2 female, and 3 female. Since you are asked for two different probabilities, you might as well get practice with making these kinds of calculations by finding the probability of each of the four possible combinations. That will also give you confidence in your methods and your calculations, if the sum of the four probabilities is 1.
# ways to choose 3 of the 14: 14C3 = 364
# ways to choose 3 male: (8C3)*(6C0) = 56*1 = 56
# ways to choose 2 male and 1 female: (8C2)(6C1) = 28*6 = 168
# ways to choose 1 male and 2 female: (8C1)(6C2) = 8*15 = 120
# ways to choose 0 male and 3 female: (8C0)(6C3) = 1*20 = 20
Observe that the sum of the different ways is the correct total of 364.
Now it's easy to answer the two questions.
(a) at least one male; i.e., 1, 2, or 3 male: (120+168+56)/364 = 344/364 = 86/91
(or (364-20)/364, since "at least one male" means NOT 0 male and 3 female)
(b) at least two male; i.e, 2 or 3 male: (168+56)/364 = 224/364 = 56/91 = 8/13
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