Question 1118588: A pharmacist wishes to select three brands of aspirin to sell in his store. He has five major brands to choose from: A, B, C, D and E. If he selects the three brands at random, what is the probability that he will select the following?
(a)
brand B
(b)
brands B and C
(c)
at least one of the two brands B and C
Answer by greenestamps(13198) (Show Source):
You can put this solution on YOUR website!
The number of ways he can choose 3 of the 5 brands is C(5,3) = 10. So the denominator of the probability fraction is 10.
The numerators are the numbers of ways of choosing the specified brands.
(a) brand B
He has to choose brand B (C(1,1)); and he can choose any 2 of the other 4 (C(4,2)). The probability is then (C(1,1)*C(4,2))/C(5,3) = (1*6)/10 = 6/10 = 3/5.
(b) brands B and C
He has to choose both brands B and C (C(2,2)); and then he can choose any 1 of the other 3 (C(3,1). You can figure out the probability for this one.
(c) at least one of the two brands B and C
The opposite of this is choosing neither B nor C and all 3 of the other brands. There is clearly only one way to do that; the probability of choosing neither B nor C is 1/10; so the probability of choosing at least one of B or C is 1-1/10 = 9/10.
Note that the number of different ways to choose 3 of the 5 brands (10) is so small that you can verify any answers you get using formal combinatorical methods by listing all ten combinations and seeing which ones satisfy the given conditions.
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