A lot of 100 bulbs contains 10 defective bulb | vedclass

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Question

(A) Let $X$ denote the number of defective bulbs. $p$ is the probability that a bulb is defective: $p = \frac{10}{100} = \frac{1}{10}$ $q = 1 - p = 1

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$\frac{7}{5}\left(\frac{9}{10}\right)^4$

Vedclass · Answer

(A) Let $X$ denote the number of defective bulbs. $p$ is the probability that a bulb is defective: $p = \frac{10}{100} = \frac{1}{10}$ $q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}$ Since the bulbs are selected from a large lot,we use the binomial distribution: $P(X = r) = { }^5 C_r \left(\frac{1}{10}ight)^r \left(\frac{9}{10}ight)^{5-r}, r = 0, 1, \dots, 5$ We need to find the probability that the store receives at most one defective bulb,i.e.,$P(X \leq 1)$: $P(X \leq 1) = P(X = 0) + P(X = 1)$ $P(X = 0) = { }^5 C_0 \left(\frac{1}{10}ight)^0 \left(\frac{9}{10}ight)^5 = \left(\frac{9}{10}ight)^5$ $P(X = 1) = { }^5 C_1 \left(\frac{1}{10}ight)^1 \left(\frac{9}{10}ight)^4 = 5 	imes \frac{1}{10} 	imes \left(\frac{9}{10}ight)^4 = \frac{1}{2} \left(\frac{9}{10}ight)^4$ $P(X \leq 1) = \left(\frac{9}{10}ight)^5 + \frac{1}{2} \left(\frac{9}{10}ight)^4 = \left(\frac{9}{10}ight)^4 \left[ \frac{9}{10} + \frac{1}{2} ight] = \left(\frac{9}{10}ight)^4 \left[ \frac{9+5}{10} ight] = \left(\frac{9}{10}ight)^4 \left( \frac{14}{10} ight) = \frac{7}{5} \left(\frac{9}{10}ight)^4$

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

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