In a box containing 100 bulbs,10 are defectiv | vedclass

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(D) The selection of bulbs from a box can be modeled as Bernoulli trials. Let $X$ denote the number of defective bulbs in a sample of $5$ bulbs.The pr

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$\left(\frac{9}{10}\right)^{5}$

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(D) The selection of bulbs from a box can be modeled as Bernoulli trials. Let $X$ denote the number of defective bulbs in a sample of $5$ bulbs.The probability of selecting a defective bulb is $p = \frac{10}{100} = \frac{1}{10}$.Therefore,the probability of selecting a non-defective bulb is $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$.Since the sample size $n = 5$ is small relative to the total population,we can treat this as a binomial distribution $X \sim B(n, p)$ with $n = 5$ and $p = \frac{1}{10}$.The probability mass function is given by $P(X = x) = ^{n}C_{x} \cdot p^{x} \cdot q^{n-x} = ^{5}C_{x} \cdot \left(\frac{1}{10}\right)^{x} \cdot \left(\frac{9}{10}\right)^{5-x}$.We need to find the probability that none of the bulbs are defective,which is $P(X = 0)$.$P(X = 0) = ^{5}C_{0} \cdot \left(\frac{1}{10}\right)^{0} \cdot \left(\frac{9}{10}\right)^{5-0}$.$P(X = 0) = 1 \cdot 1 \cdot \left(\frac{9}{10}\right)^{5} = \left(\frac{9}{10}\right)^{5}$.The correct answer is $D$.

In a box containing $100$ bulbs,$10$ are defective. The probability that out of a sample of $5$ bulbs,none is defective is

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