On an average,if one out of 100 electric bulb | vedclass

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Question

(A) Let $X$ be the number of defective bulbs in a consignment of $n = 600$ bulbs. The probability of a bulb being defective is $p = \frac{1}{100} = 0.

Vedclass · Accepted Answer

$1-7 e^{-6}$

Vedclass · Answer

(A) Let $X$ be the number of defective bulbs in a consignment of $n = 600$ bulbs. The probability of a bulb being defective is $p = \frac{1}{100} = 0.01$. Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $\lambda = np = 600 	imes 0.01 = 6$. The probability of having $k$ defective bulbs is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-6} 6^k}{k!}$. We need to find the probability of at least two defective bulbs,which is $P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$. Calculating the probabilities: $P(X = 0) = \frac{e^{-6} 6^0}{0!} = e^{-6}$. $P(X = 1) = \frac{e^{-6} 6^1}{1!} = 6e^{-6}$. Therefore,$P(X \ge 2) = 1 - (e^{-6} + 6e^{-6}) = 1 - 7e^{-6}$. The correct option is $A$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

$x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$0.2$	$0.5$	$0.15$	$0.25$	$0.1$

On an average,if one out of $100$ electric bulbs produced by a company is found to be defective,then the probability that there are at least two defective bulbs in a consignment of $600$ bulbs is:

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