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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

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

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

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X=x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

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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