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# If X is a Poisson variate such that $P(2)=9\: P(4)+90\: P(6)$. then mean of X is

$\begin {array} {1 1} (A)\;\pm\: 1 & \quad (B)\;\pm\: 2 \\ (C)\;\pm\: 3 & \quad (D)\;None\: of \: these \end {array}$

Using the given relation we have
$\Large\frac{e^{-m}m^2}{2!} = \Large\frac{9e^{-m}m^4}{4!} +\Large\frac{90e^{-m}m^6}{6!}$
$= m^4+3m^2-4=0$
$(m^2-1)(m^2+4)=0$
$m = \pm 1$
$m^2=-4$ ( not feasible )
Hence $m = \pm 1$
Ans : (A)