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(A) Let $X$ be the number of defective units in a sample of $n = 5$ units.
The probability of a unit being defective is $p = \frac{1}{8}$, and the pr

Vedclass · Accepted Answer

$\frac{7203}{8192}$

Vedclass · Answer

(A) Let $X$ be the number of defective units in a sample of $n = 5$ units.
The probability of a unit being defective is $p = \frac{1}{8}$, and the probability of a unit being non-defective is $q = 1 - p = \frac{7}{8}$.
Since the trials are independent, $X$ follows a binomial distribution $B(n, p) = B(5, \frac{1}{8})$.
The probability of having at most one defective unit is $P(X \le 1) = P(X = 0) + P(X = 1)$.
Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k q^{n-k}$:
$P(X = 0) = \binom{5}{0} (\frac{1}{8})^0 (\frac{7}{8})^5 = 1 	imes 1 	imes (\frac{7}{8})^5 = \frac{16807}{32768}$.
$P(X = 1) = \binom{5}{1} (\frac{1}{8})^1 (\frac{7}{8})^4 = 5 	imes \frac{1}{8} 	imes \frac{2401}{4096} = \frac{12005}{32768}$.
$P(X \le 1) = \frac{16807 + 12005}{32768} = \frac{28812}{32768} = \frac{7203}{8192}$.
None of the given options match the calculated result $\frac{7203}{8192}$.

Among every $8$ units of a product, one is likely to be defective. If a consumer has ordered $5$ units of that product, then the probability that at most one unit is defective among them is

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