If 20 % of the bolts produced by a machine ar | vedclass

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(C) Let $X$ be the number of defective bolts in a sample of $n=4$. The probability of a bolt being defective is $p = 20\% = 0.2 = \frac{1}{5}$.Thus, t

Vedclass · Accepted Answer

$0.8192$

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(C) Let $X$ be the number of defective bolts in a sample of $n=4$. The probability of a bolt being defective is $p = 20\% = 0.2 = \frac{1}{5}$.Thus, the probability of a bolt being non-defective is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.Since the selection is random, $X$ follows a binomial distribution $B(n, p)$ with $n=4$ and $p=\frac{1}{5}$.The probability mass function is given by $P(X=k) = {}^nC_k p^k q^{n-k}$.We need to find the probability that less than $2$ bolts are defective, i.e.,$P(X $P(X=0) = {}^4C_0 \left(\frac{1}{5}ight)^0 \left(\frac{4}{5}ight)^4 = 1 	imes 1 	imes \frac{256}{625} = \frac{256}{625}$.$P(X=1) = {}^4C_1 \left(\frac{1}{5}ight)^1 \left(\frac{4}{5}ight)^3 = 4 	imes \frac{1}{5} 	imes \frac{64}{125} = \frac{256}{625}$.Therefore, $P(X < 2) = \frac{256}{625} + \frac{256}{625} = \frac{512}{625} = 0.8192$.

If $20 \%$ of the bolts produced by a machine are defective, then the probability that out of $4$ bolts chosen at random, less than $2$ bolts will be defective, is

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