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(A) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k !}$.Here,$\lambda$ is the mean numb

Vedclass · Accepted Answer

$\sum_{k=3}^{\infty} \frac{e^{-0.2} (0.2)^k}{k !}$

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(A) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k !}$.Here,$\lambda$ is the mean number of accidents per day.Given that $10$ accidents occur in $50$ days,the average number of accidents per day is $\lambda = \frac{10}{50} = 0.2$.We need to find the probability that three or more accidents occur in a day,which is $P(X \geq 3)$.$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) + \dots$This can be written as $\sum_{k=3}^{\infty} \frac{e^{-\lambda} \lambda^k}{k !}$ with $\lambda = 0.2$.Thus,the correct expression is $\sum_{k=3}^{\infty} \frac{e^{-0.2} (0.2)^k}{k !}$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$K$	$2K$	$3K$	$4K$	$5K$	$6K$

In a city,$10$ accidents take place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that three or more accidents occur in a day is:

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