SOLUTION: A computer dealer had 20 computers of which 3 are defective. A. If she selects a computer at random what is the probability that it is defective? B. If she selects computers at r

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Question 247835: A computer dealer had 20 computers of which 3 are defective.
A. If she selects a computer at random what is the probability that it is defective?
B. If she selects computers at random, what is the probability that one of them is defective?
C. If she selects 3 computers at random what is the probability that all of them are defective?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
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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