Probability that binomial random variable is even Suppose $X$ is binomial $B(n,p)$. How can I find the probability that $X$ is even ?
I know $$P(X = k ) = \frac{ n!}{(n-k)!k!} p^k(1-p)^{n-k} $$
where $X=1,....,n$. Are they just asking to find $P(X = 2m )$ for some $m > 0$ ?

Here is a perhaps more direct solution. We will use one "trick;" the following function is equal to $1$ when $k$ is an even integer, and equal to $0$ when $k$ is an odd integer:
$$
\frac{(-1)^k+1}2
$$
Therefore,
$$
P(X\text{ is even})=\sum_{k=0}^n \frac{(-1)^k+1}{2}\cdot P(X=k)=\frac12\sum_{k=0}^nP(X=k)+\frac12\sum_{k=0}^n (-1)^k P(X=k)
$$
The first summation, $\sum_{k=0}^nP(X=k)$, obviously equals one. For the second, if you plug in $\binom{n}{k}p^k(1-p)^{n-k}$ for $P(X=k)$, you will find that this summation can be simplified using the binomial theorem.
