SOLUTION: A computer retail store has 9 personal computers in stock. A buyer wants to purchase 4 of them. Unknown to either the retail store or the buyer, 4 of the computers in stock have d

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Question 1145993: A computer retail store has 9 personal computers in stock. A buyer wants to purchase 4 of them. Unknown to either the retail store or the buyer, 4 of the computers in stock have defective hard drives. Assume that the computers are selected at random.
(a) In how many different ways can the 4 computers be chosen?
(b) What is the probability that exactly one of the computers will be defective?
(c) What is the probability that at least one of the computers selected is defective?
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
(a) In how many different ways can the 4 computers be chosen?

9C4 = 9%21%2F%284%21%2A5%21%29 = 126

Since 4 out of the 9 computers are defective, this means there is a 0.4444 probability a randomly chosen computer is defective. (You will need to know this to answer the next two questions.)

(b) What is the probability that exactly one of the computers will be defective?

%280.4444%29%5E1+%2A+%280.5556%29%5E3+%2A+%284%21%2F%281%21%2A3%21%29%29 = 0.4444 * 0.1715 * 4 = 0.3049

(c) What is the probability that at least one of the computers selected is defective?

To do this, you compute the probability that NONE of the four computers is defective, then subtract this result from 1:

1+-+%280.5556%29%5E4 = 1 - 0.0953 = 0.9047