Answer: 0.1

## Explanation For a Poisson random variable, the probability mass function is: \[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\] Given: - \[P(X = 1) = 0.09 = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda\] ...(i) - \[P(X = 2) = 0.0045 = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^2}{2} = 0.0045\] From equation (ii): \[\frac{e^{-\lambda} \lambda^2}{2} = 0.0045\] \[e^{-\lambda} \lambda^2 = 0.009\] ...(ii) Now, divide equation (ii) by equation (i): \[\frac{e^{-\lambda} \lambda^2}{e^{-\lambda} \lambda} = \frac{0.009}{0.09}\] \[\lambda = 0.1\] For a Poisson distribution, the variance equals the mean: \[\text{Var}(X) = \lambda = 0.1\] Therefore, the correct answer is **D. 0.1**

Author: Tanishq Prabhu