In a city,it is found that 10 accidents took | vedclass

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Question

(B) Let $X$ be the random variable representing the number of accidents in a day. $X \sim 	ext{Poisson}(\lambda)$.Given that $10$ accidents occurred

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$1-(1.22) e^{-0.2}$

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(B) Let $X$ be the random variable representing the number of accidents in a day. $X \sim 	ext{Poisson}(\lambda)$.Given that $10$ accidents occurred in $50$ days,the mean rate $\lambda$ per day is $\lambda = \frac{10}{50} = 0.2$.The probability of $X$ accidents in a day is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.We need to find $P(X \geq 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]$.$P(X \geq 3) = 1 - \left[ \frac{e^{-0.2} (0.2)^0}{0!} + \frac{e^{-0.2} (0.2)^1}{1!} + \frac{e^{-0.2} (0.2)^2}{2!} ight]$.$P(X \geq 3) = 1 - e^{-0.2} \left[ 1 + 0.2 + \frac{0.04}{2} ight]$.$P(X \geq 3) = 1 - e^{-0.2} [1 + 0.2 + 0.02] = 1 - 1.22 e^{-0.2}$.

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

In a city,it is found that $10$ accidents took place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that there will be $3$ or more accidents in a day in that city is

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