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(D) Given,the mean of the Poisson distribution is $\lambda = \frac{1}{3}$.For a Poisson distribution,the probability mass function is given by $P(X=k)

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$6 : 1$

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(D) Given,the mean of the Poisson distribution is $\lambda = \frac{1}{3}$.For a Poisson distribution,the probability mass function is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$.For $k=1$,$P(X=1) = \frac{e^{-1/3} (1/3)^1}{1!} = \frac{1}{3} e^{-1/3}$.For $k=2$,$P(X=2) = \frac{e^{-1/3} (1/3)^2}{2!} = \frac{1}{2} 	imes \frac{1}{9} e^{-1/3} = \frac{1}{18} e^{-1/3}$.Now,the ratio $P(X=1) : P(X=2) = \frac{\frac{1}{3} e^{-1/3}}{\frac{1}{18} e^{-1/3}} = \frac{1/3}{1/18} = \frac{18}{3} = 6$.Thus,the ratio is $6 : 1$.

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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