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There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection?

can we do it by both hypergeometric as well as by binomial distribution?
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Binomial used for------------>

a) In such trial we have either success or failure ..

b) Sampling should be from infinite population(here it is referred to no of caculator)

c) Sampling from finite population but with replacement

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here B and C condition is false because its finite . and default we take without replacement so here Hypergeometric is apply

we get,

$=\dfrac{{^2}C{_1}\times {^{23}}C{_4}}{{^{25}}C{_5}}.$
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Q. - > can we do it by both hyper geometric as well as by binomial distribution?

No.

Many people have given correct answer for this question. Now just mentioning another approach to get desired answer -->

We have to select  5 items in which 4 are of one type(non-defective ) and 1 is of another type (defective).

$\Rightarrow \frac{5!}{4!\,1!} $ * (probability of selecting these items) $= \frac{5!}{4!\,1!} * \left ( \frac{2}{25}*\frac{23}{24}*\frac{22}{23}*\frac{21}{22}*\frac{20}{21} \right )  =  \frac{\binom{2}{1}*\binom{23}{4}}{\binom{25}{5}}$
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