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If $X$ is a Poisson variate and $P(X=1)=2P(X=2)$ then $P(X=3)=$

\[(a)\;\frac{e^{-1}}{6} \quad (b)\;\frac{e^{-2}}{2} \quad (c)\;\frac{e^{-1}}{2} \quad (d)\;\frac{e^{-1}}{3} \]

1 Answer

$(a)\;\frac{e^{-1}}{6}$
answered Nov 7, 2013 by pady_1
 
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