Gate-2006, CE

Question

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?

Comments

@prerna . CHeck this link: https://gateoverflow.in/88215/bulbs-out-sample-bulbs-manufactured-company-are-defective

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 How ans is 1/3?

 ^{value of   (2}c_1* ²³c₄)/ ²⁵c_{5 = 1 / 60} 

Plz check it.

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This question cannot be solved by Binomial Distribution, bcoz there are not Independent trials.A que can be solved by Binomial distribution if and only if there are n- Independent trials and in every trials prob of success is must be the same p that is why we call as Bin(n, p).

Answer 1

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,

$=\dfrac{{^2}C{_1}\times {^{23}}C{_4}}{{^{25}}C{_5}}.$

Comments

elaborate it more,please?

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This question cannot be solved by Binomial Distribution, bcoz there are not Independent trials.A que can be solved by Binomial distribution if and only if there are n- Independent trials and in every trials prob of success is must be the same p that is why we call as Bin(n, p).

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Here the parameters of hypergeometric distribution is as follows:

K=2  acceptable items

N-K=23 defective items(total items, N=25)

n =5  is the size of items of same type to be chosen without replacement

and k=1.

Answer 2

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