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If the mean of a poisson distribution is $\frac{1}{3}$, then the ratio $P(X=1): P(X=2)$ is
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$6: 1$
Given, Mean of Poisson distribution $=1 / 3$
For a Poisson distribution, the mean is equal to the parameter. Hence, the parameter is $1 / 3$.
$\therefore \quad P(X=1)=\frac{e^{-1 / 3}(1 / 3)^1}{1 !}$
$=e^{-1 / 3}(1 / 3)$
$P(X=2)=\frac{e^{-1 / 3}\left(\frac{1}{3}\right)^2}{2 !}$
$\therefore \quad P(X=1): P(X=2)=\frac{e^{-1 / 3}\left(\frac{1}{3}\right)}{\frac{e^{-1 / 3}\left(\frac{1}{3}\right)^2}{2 !}}$
$=\frac{2}{1 / 3}=\frac{6}{1}=6: 1$
For a Poisson distribution, the mean is equal to the parameter. Hence, the parameter is $1 / 3$.
$\therefore \quad P(X=1)=\frac{e^{-1 / 3}(1 / 3)^1}{1 !}$
$=e^{-1 / 3}(1 / 3)$
$P(X=2)=\frac{e^{-1 / 3}\left(\frac{1}{3}\right)^2}{2 !}$
$\therefore \quad P(X=1): P(X=2)=\frac{e^{-1 / 3}\left(\frac{1}{3}\right)}{\frac{e^{-1 / 3}\left(\frac{1}{3}\right)^2}{2 !}}$
$=\frac{2}{1 / 3}=\frac{6}{1}=6: 1$
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