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(A) Let $X$ be the number of defective items in a sample of $n = 8$. The probability of a defective item is $p = 5\% = \frac{5}{100} = \frac{1}{20}$.T

Vedclass · Accepted Answer

$\frac{27}{20} \left( \frac{19}{20} \right)^7$

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(A) Let $X$ be the number of defective items in a sample of $n = 8$. The probability of a defective item is $p = 5\% = \frac{5}{100} = \frac{1}{20}$.Thus,the probability of a non-defective item is $q = 1 - p = \frac{19}{20}$.We use the binomial distribution formula: $P(X = k) = {}^nC_k p^k q^{n-k}$.We need to find the probability that there are less than $2$ defective items,which is $P(X $P(X = 0) = {}^8C_0 \left( \frac{1}{20} ight)^0 \left( \frac{19}{20} ight)^8 = \left( \frac{19}{20} ight)^8$.$P(X = 1) = {}^8C_1 \left( \frac{1}{20} ight)^1 \left( \frac{19}{20} ight)^7 = 8 	imes \frac{1}{20} 	imes \left( \frac{19}{20} ight)^7 = \frac{2}{5} \left( \frac{19}{20} ight)^7$.$P(X < 2) = \left( \frac{19}{20} ight)^8 + \frac{2}{5} \left( \frac{19}{20} ight)^7 = \left( \frac{19}{20} ight)^7 \left( \frac{19}{20} + \frac{2}{5} ight) = \left( \frac{19}{20} ight)^7 \left( \frac{19 + 8}{20} ight) = \frac{27}{20} \left( \frac{19}{20} ight)^7$.

The items produced by a firm are supposed to contain $5\%$ defective items. The probability that a sample of $8$ items will contain less than $2$ defective items is:

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