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Question

(A) Let $X$ denote the number of defective items in a sample of $10$ items. Since the items are drawn from a large bulk,the trials are Bernoulli trial

Vedclass · Accepted Answer

$\left(\frac{29}{20}\right) \cdot \left(\frac{19}{20}\right)^{9}$

Vedclass · Answer

(A) Let $X$ denote the number of defective items in a sample of $10$ items. Since the items are drawn from a large bulk,the trials are Bernoulli trials.Given $p = \frac{5}{100} = \frac{1}{20}$ and $q = 1 - p = \frac{19}{20}$.$X$ follows a binomial distribution with $n = 10$ and $p = \frac{1}{20}$.The probability mass function is $P(X = x) = ^{10}C_{x} \left(\frac{19}{20}\right)^{10-x} \left(\frac{1}{20}\right)^{x}$.We need to find $P(X \leq 1) = P(X = 0) + P(X = 1)$.$P(X = 0) = ^{10}C_{0} \left(\frac{19}{20}\right)^{10} \left(\frac{1}{20}\right)^{0} = \left(\frac{19}{20}\right)^{10}$.$P(X = 1) = ^{10}C_{1} \left(\frac{19}{20}\right)^{9} \left(\frac{1}{20}\right)^{1} = 10 \cdot \left(\frac{19}{20}\right)^{9} \cdot \frac{1}{20} = \frac{1}{2} \cdot \left(\frac{19}{20}\right)^{9}$.$P(X \leq 1) = \left(\frac{19}{20}\right)^{10} + \frac{1}{2} \cdot \left(\frac{19}{20}\right)^{9} = \left(\frac{19}{20}\right)^{9} \left[ \frac{19}{20} + \frac{10}{20} \right] = \left(\frac{29}{20}\right) \cdot \left(\frac{19}{20}\right)^{9}$.

There are $5 \%$ defective items in a large bulk of items. What is the probability that a sample of $10$ items will include not more than one defective item?

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