Explanation
P(X=2)= \cfrac {2}{3}P(X=1)
P(2; \mu) = \cfrac { { e }^{-\mu}{\mu}^{2}}{2!} = \cfrac {2}{3} P(1; \mu) = \cfrac {2}{3} \cfrac { { e }^{-\mu}{\mu}^{1}}{1!}
{ e }^{-\mu} \mu (\cfrac {\mu}{2} - \cfrac {2}{3})=0
either \mu = 0\ or \mu = \cfrac{4}{3}
but here \mu = \cfrac{4}{3}P(X=3) = \cfrac { { e }^{-\mu}{\mu}^{3}}{3!} = \cfrac { { e }^{ \frac { -4 }{ 3 } }{( \frac { 4 }{ 3 } })^{ 3 } }{ 6 } = \cfrac { 64 }{ 162 } { e }^{ \frac{-4}{3} }
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