Let X∼Poisson(λ)X \sim \text{Poisson}(\lambda)X∼Poisson(λ). What is the probability that XXX takes an even value?
1+e−2λ2\frac{1 + e^{-2\lambda}}{2}21+e−2λ
1−e−2λ2\frac{1 - e^{-2\lambda}}{2}21−e−2λ
e−λsinh(λ)e^{-\lambda} \sinh(\lambda)e−λsinh(λ)
12\frac{1}{2}21