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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 $

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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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