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Doss [256]
3 years ago
15

A shipment of 12 microwave ovens contains 4 defective units. A restaurant buys three of these units. What is the probability of

the restaurant buying at least two nondefective​ units?
Mathematics
1 answer:
hammer [34]3 years ago
5 0
The <u>correct answer</u> is:

20/27 = 0.7407.

Explanation:

To find the probability that the restaurant buys at least 2 non-defective units, we find the probability that they purchase either 2 or 3 non-defective units:

P(r) = _nC_r(p)^r(1-p)^{n-r}&#10;

They purchase <u>3 units total</u>, so this is <u>n</u>.
We want the <u>probability of either 2 or 3</u> non-defective units; this means <u>r is 2 and r is 3</u>.
In this case we want the probability that the units are non-defective.  Since there are 4 defective units, this means there are 12-4=8 non-defective units; this makes the <u>probability of a non-defective unit 8/12 = 2/3.  This is p</u>.

Using these, we have:
P(X=2 or X=3)=_3C_2(\frac{2}{3})^2(1-\frac{2}{3})^{3-2}+_3C_3(\frac{2}{3})^3(1-\frac{2}{3})^{3-3}&#10;\\&#10;\\=\frac{3!}{2!1!}(\frac{2}{3})^2(\frac{1}{3})^1+\frac{3!}{3!0!}(\frac{2}{3})^3(\frac{2}{3})^0&#10;\\&#10;\\=3(\frac{4}{9})(\frac{1}{3})+1(\frac{8}{27})(1)&#10;\\&#10;\\=\frac{12}{27}+\frac{8}{27}=\frac{20}{27}=0.7407
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