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A packet of 20 batteries is known to include 4 batteries that are defective. If 8 batteries are randomly chosen and tested, the probability that finding among them not more than 1 defective is

Ans: 0.5033

Solution provided:

 

How can we apply Binomial distribution here? The batteries are chosen without replacement right? Then how can the probability of being defective or not defective remain same?

Shouldn’t it be

$\frac{ \binom{16}{8} \times \binom{4}{0}}{ \binom{20}{8}} +\frac{ \binom{16}{7} \times \binom{4}{1}}{ \binom{20}{8}}=0.465 $

in Quantitative Aptitude
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0

 we are choosing 8 batteries collectively not one by one that's why probability will be same .

0

@Prateek Raghuvanshi

Even if we pick them up collectively the same formula should be applied isnt't it?

If we pick them up with replacement then we have to follow ME's solution..

https://www.futureaccountant.com/probability/problems-solutions/drawing-picking-selecting-two-three-more-balls-box-bag-urn-07.php#.XCyWB81S_IU

This is what I  knew.. correct me if I'm wrong

0

 there are 20 batteries in the packet and in which 4 are defective but we don't know which one is defective so probability of defective will be 4/20 and for non defective 4/5 .when we are choosing 8 batteries we don't  know which one is defective or non defective  so for every battery probability will be either of defective or of non defective ,which will be same for all batteries (Bernoulli trail) .

let me know you are getting my point or not.

0

 according to me your answer is correct. @Arjun sir please check

0

@Prateek Raghuvanshi

We use Bernoulli's trial when the success/failure probability remain constant i.e. in case of rolling a die or tossing coin in each trial..

0

 according to me  here also probability is constant of defective and non defective  .

0

why can't we use hypergeometric distribution here?? @Arjun sir please check

$(\binom{4}{1}*\binom{16}{7} + \binom{4}{0}*\binom{16}{8}) / \binom{20}{8}$

1 Answer

2 votes
 
Best answer
Probability of getting $0$ defective $ = \frac{{}^{16}C_8}{{}^{20}C_8} = \frac{16!12!}{20!8!} = \frac{9.10.11.12}{17.18.19.20} = \frac{11.3}{17.19}=0.1021$

Probability of getting $1$ defective $ = \frac{{}^{16}C_7. {}^4C_1}{{}^{20}C_8} = \frac{16!12!8.4}{20!9!} = \frac{10.11.12.8.4}{17.18.19.20} = \frac{11.12.8}{17.9.19} =0.3632 $

So, probability of not more than one defective $ = 0.102+0.3632 = 0.4654$

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