The aim is to compute P(X>0)P(X>0)P(X>0). We remind that for the Poisson distribution with the mean equal to 0.40.40.4 we have
P(X=k)=(0.4)kk!e−0.4.P(X=k)= \frac{(0.4)^{k}}{k!}e^{-0.4}.P(X=k)=k!(0.4)ke−0.4.
Thus, using the Taylor series for the exponent, we have
P(X>0)=∑k=1+∞(0.4)kk!e−0.4=e−0.4(e0.4−1)=0.3297P(X>0)=\sum_{k=1}^{+\infty} \frac{(0.4)^{k}}{k!}e^{-0.4}=e^{-0.4}(e^{0.4}-1)=0.3297P(X>0)=∑k=1+∞k!(0.4)ke−0.4=e−0.4(e0.4−1)=0.3297
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