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(A) The number of accidents follows a Poisson distribution with parameter $\lambda = 3$. The probability mass function of a Poisson random variable $X

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$\frac{1}{e^3}$

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(A) The number of accidents follows a Poisson distribution with parameter $\lambda = 3$. The probability mass function of a Poisson random variable $X$ is given by $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$,where $x = 0, 1, 2, \dots$. We want to find the probability that no accidents occur today,which corresponds to $P(X = 0)$. Substituting $\lambda = 3$ and $x = 0$ into the formula: $P(X = 0) = \frac{3^0 e^{-3}}{0!}$. Since $3^0 = 1$ and $0! = 1$,we get: $P(X = 0) = \frac{1 	imes e^{-3}}{1} = e^{-3} = \frac{1}{e^3}$. Thus,the probability that no accidents occur today is $\frac{1}{e^3}$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$F(X=x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

Suppose the number of accidents occurring on a highway in each day follows a Poisson random variable with parameter $3$. Then,what is the probability that no accidents occur today?

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