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OpenStudy (anonymous):
If 5 of a company's 12 delivery trucks do not meet emission standards and 4 of the twelve trucks are randomly pick for inspection, what is the probability that none of them meet emission standards
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OpenStudy (anonymous):
two ways to proceed. simplest way is to say first one doesn't, then second one doesn't , then third on doesn't then fourth one doesn't and compute \[\frac{5}{12}\times \frac{4}{11}\times \frac{3}{10}\times \frac{2}{9}\]
OpenStudy (anonymous):
second way is to say the number of ways to pick 4 out of 12 is \(\binom{12}{4}\) and the number of ways to pick 4 out of the 5 that don't meet the standards is \(\binom{5}{4}=5\)
OpenStudy (anonymous):
and then compute \[\frac{\dbinom{5}{4}}{\dbinom{12}{4}}=\frac{5}{792}\]
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