Question 1145993
<i>(a) In how many different ways can the 4 computers be chosen?</i>
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9C4 = {{{9!/(4!*5!)}}} = 126
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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.)
<pr>
<i>(b) What is the probability that exactly one of the computers will be defective?</i>
<pr>
{{{(0.4444)^1 * (0.5556)^3 * (4!/(1!*3!))}}} = 0.4444 * 0.1715 * 4 = 0.3049
<pr>
<i>(c)  What is the probability that at least one of the computers selected is defective?</i>
<pr>
To do this, you compute the probability that NONE of the four computers is defective, then subtract this result from 1:
<pr>
{{{1 - (0.5556)^4}}} = 1 - 0.0953 = 0.9047