If X is a Poisson random variate with mean 3, | vedclass

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(B) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}$.Given $\lambda = 3$,we nee

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$\frac{99}{8 e^3}$

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(B) For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}$.Given $\lambda = 3$,we need to find $P(|X-3| The inequality $|X-3| Since $X$ is a discrete random variable taking non-negative integer values,the possible values for $X$ are $2, 3, 4$.Thus,$P(|X-3| Substituting the values into the formula:$P(X=2) = \frac{3^2 \cdot e^{-3}}{2!} = \frac{9}{2} e^{-3}$$P(X=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27}{6} e^{-3} = \frac{9}{2} e^{-3}$$P(X=4) = \frac{3^4 \cdot e^{-3}}{4!} = \frac{81}{24} e^{-3} = \frac{27}{8} e^{-3}$Summing these probabilities:$P(|X-3| < 2) = e^{-3} \left( \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) = e^{-3} \left( 9 + \frac{27}{8} \right) = e^{-3} \left( \frac{72+27}{8} \right) = \frac{99}{8 e^3}$.

If $X$ is a Poisson random variate with mean $3$,then $P(|X-3| < 2) =$

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