A lot of 100 bulbs contains 10 defective bulb | vedclass

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(B) Let $p$ be the probability of selecting a defective bulb. Given $p = \frac{10}{100} = 0.1$ and $q = 1 - p = 0.9$.We select $n = 5$ bulbs. Let $X$

Vedclass · Accepted Answer

$0.91854$

Vedclass · Answer

(B) Let $p$ be the probability of selecting a defective bulb. Given $p = \frac{10}{100} = 0.1$ and $q = 1 - p = 0.9$.We select $n = 5$ bulbs. Let $X$ be the number of defective bulbs.We need to find $P(X \le 1) = P(X = 0) + P(X = 1)$.Using the binomial distribution formula $P(X = k) = {}^nC_k p^k q^{n-k}$:$P(X = 0) = {}^5C_0 (0.1)^0 (0.9)^5 = 1 	imes 1 	imes (0.9)^5 = 0.59049$.$P(X = 1) = {}^5C_1 (0.1)^1 (0.9)^4 = 5 	imes 0.1 	imes 0.6561 = 0.5 	imes 0.6561 = 0.32805$.Therefore,$P(X \le 1) = 0.59049 + 0.32805 = 0.91854$.

$A$ lot of $100$ bulbs contains $10$ defective bulbs. Five bulbs are selected at random from the lot and sent to a retail store. The probability that the store will receive at most one defective bulb is:

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