<PRE><B>Question 247835</B>
A computer dealer had 20 computers of which 3 are defective.
Binomial with n=20 ; p=3/20 ; x varies
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A. If she selects a computer at random what is the probability that it is defective?
P(one selected defective) = 3/20
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B. If she selects computers at random, what is the probability that one of them is defective?
That depends on how many computers she selects.
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C. If she selects 3 computers at random what is the probability that all of them are defective?
P(all of 3 are defective) 
# of ways to select 3 defective: 1
# of ways to select 3 without restriction: 20C3 = 1140
P(3 of 3 are defective) = 1/1140
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Another way to work this is:
(3/20)(2/19)(1/18) = 1/1140
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Cheers,
Stan H.
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