If a random variable X follows Poisson\'s distribution such that P (X…

The correct answer is Option (1) → 1

For a Poisson distribution with mean λ, \( P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \).

    \( \displaystyle \frac{e^{-\lambda}\lambda^2}{2}=9\cdot\frac{e^{-\lambda}\lambda^4}{24}+90\cdot\frac{e^{-\lambda}\lambda^6}{720} \)

Cancelling \( e^{-\lambda} \) and simplifying the coefficients:

    \( \displaystyle \frac{\lambda^2}{2}=\frac{3\lambda^4}{8}+\frac{\lambda^6}{8} \)

Multiplying by \( \frac{8}{\lambda^2} \) (λ > 0):

    \( \displaystyle 4=3\lambda^2+\lambda^4 \)
    \( \displaystyle \lambda^4+3\lambda^2-4=0 \)
    \( \displaystyle (\lambda^2+4)(\lambda^2-1)=0 \)

Since \( \lambda^2+4\gt 0 \), \( \lambda^2=1 \), so λ = 1. The mean of X is 1.

Shortcut: Put λ = 1 in \( 4=3\lambda^2+\lambda^4 \): 3 + 1 = 4. For λ = 2 the right side is 28, for λ = 1/2 it is 13/16, and for λ = 1/4 it is less still.