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A random variable $X$ has Poisson distribution with mean 2. Then $P(X>1.5)$ equals
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Verified Answer
The correct answer is:
$1-\frac{3}{\mathrm{e}^2}$
$1-\frac{3}{\mathrm{e}^2}$
$$
\begin{aligned}
& P(x=k)=e^{-\lambda} \frac{\lambda^k}{k !} \\
& P(x \geq 2)=1-P(x=0)-P(x=1) \\
& =1-e^{-\lambda}-e^{-\lambda}\left(\frac{\lambda}{1 !}\right) \\
& =1-\frac{3}{e^2} .
\end{aligned}
$$
\begin{aligned}
& P(x=k)=e^{-\lambda} \frac{\lambda^k}{k !} \\
& P(x \geq 2)=1-P(x=0)-P(x=1) \\
& =1-e^{-\lambda}-e^{-\lambda}\left(\frac{\lambda}{1 !}\right) \\
& =1-\frac{3}{e^2} .
\end{aligned}
$$
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