Probability and Statistics

For a Poisson distribution, $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$. $P(X = 1) = \frac{e^{-\lambda} \lambda^1}{1!} = \lambda e^{-\lambda}$ $P(X = 2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{\lambda^2 e^{-\lambda}}{2}$ Equating the two: $\lambda e^{-\lambda} = \frac{\lambda^2 e^{-\lambda}}{2}$ Since $\lambda > 0$ and $e^{-\lambda} \ne 0$: $1 = \frac{\lambda}{2} \implies \lambda = 2.0$