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

The correct answer is Option (4) → 2

For a Poisson distribution with parameter \(\lambda\),

    \( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \)

Substituting in the given relation and cancelling \(e^{-\lambda}\):

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

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

Multiplying by 8 and dividing by \(\lambda^2\) (\(\lambda \gt 0\)):

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

Since \(\lambda^2 + 4 \gt 0\), \(\lambda^2 = 1\), so \(\lambda = 1\).

In a Poisson distribution the mean and variance are both \(\lambda\):

    \(\text{Mean} + \text{Variance} = 2\lambda = 2\)