A box has 100 pens of which 10 are defective. | vedclass

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(C) Given that,the probability of selecting a defective pen is $p = \frac{10}{100} = \frac{1}{10}$.So,the probability of selecting a non-defective pen

Vedclass · Accepted Answer

$ \left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4} $

Vedclass · Answer

(C) Given that,the probability of selecting a defective pen is $p = \frac{10}{100} = \frac{1}{10}$.So,the probability of selecting a non-defective pen is $q = 1 - p = \frac{9}{10}$.Since the pens are drawn with replacement,this follows a binomial distribution with $n = 5$ trials.We need to find the probability that at most one pen is defective,i.e.,$P(X \leq 1) = P(X = 0) + P(X = 1)$.Using the binomial formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:$P(X = 0) = {}^{5}C_{0} \left(\frac{1}{10}ight)^{0} \left(\frac{9}{10}ight)^{5} = \left(\frac{9}{10}ight)^{5}$.$P(X = 1) = {}^{5}C_{1} \left(\frac{1}{10}ight)^{1} \left(\frac{9}{10}ight)^{4} = 5 	imes \frac{1}{10} 	imes \left(\frac{9}{10}ight)^{4} = \frac{1}{2} \left(\frac{9}{10}ight)^{4}$.Therefore,$P(X \leq 1) = \left(\frac{9}{10}ight)^{5} + \frac{1}{2} \left(\frac{9}{10}ight)^{4}$.

$A$ box has $100$ pens of which $10$ are defective. The probability that out of a sample of $5$ pens drawn one by one with replacement,at most one is defective is:

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