If a random variable X follows Poisson's distribution such that $P(X = 2) = 9 P(X = 4)+90 P(X = 6)$, then sum of the mean and variance of X is

2

The correct answer is Option (4) → 2 **

For a Poisson distribution with parameter $\lambda$:

$P(X=k)=e^{-\lambda}\frac{\lambda^{k}}{k!}$

Given:

$P(2)=9P(4)+90P(6)$

Substitute the formula:

$e^{-\lambda}\frac{\lambda^{2}}{2} = 9e^{-\lambda}\frac{\lambda^{4}}{24} + 90e^{-\lambda}\frac{\lambda^{6}}{720}$

Cancel $e^{-\lambda}$:

$\frac{\lambda^{2}}{2} = \frac{9\lambda^{4}}{24} + \frac{90\lambda^{6}}{720}$

$4\lambda^{2} = 3\lambda^{4} + \lambda^{6}$

Rearrange:

$\lambda^{6} + 3\lambda^{4} - 4\lambda^{2} = 0$

Factor:

$\lambda^{2}(\lambda^{4} + 3\lambda^{2} - 4)=0$

Let $t=\lambda^{2}$:

$t^{2} + 3t - 4 = 0$

$(t-1)(t+4)=0$

$t=1$ (positive root)

$\lambda^{2}=1 \Rightarrow \lambda=1$

For Poisson distribution:

Mean = $\lambda$

Variance = $\lambda$

Sum of mean and variance = $2\lambda = 2$

The required sum is $2$.