Probability that an integer number having Poisson distribution is even
Solution 1:
Hint: $$ e^x + e^{-x} = \sum_{n=0}^\infty \frac{x^n}{n!} + \sum_{n=0}^\infty \frac{(-x)^n}{n!} = 2\cdot\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!} $$
Solution 2:
Let $\displaystyle f(\lambda)= \sum_{k\ge0}\frac{\lambda^{2k}}{(2k)!}$.
Then $f''(\lambda) = f(\lambda)$. Solving this differential equation and applying the initial conditions $f(0)=1$ and $f'(0)=0$, we get $$ f(\lambda) = \frac{e^\lambda+e^{-\lambda}}{2} = \cosh(\lambda). $$
So $\Pr(\text{even}) = \dfrac{1+e^{-2\lambda}}{2}$.