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 $m > 0$ ?

I claim that the answer is $\frac{1}{2}(1+(1-2p)^n)$.
To see why, first of we have to realise what the binomial distribution actually stands for. If an event happens with probability $p$, and we make $n$ trials, then $P(X=k)$ as you defined tells us the chance that we have $k$ "successes", i.e. the chance that the event happens exactly $k$ times. We want to determine the probability that in a process of $n$ trials with success probability $p$, we have an even number of successes. Now lets say that we have performed $n-1$ trials. If we then have an odd number of successes, we will need the last trial to be a success in order for the total number of successes to be even. If the number of successes after $n-1$ trials is already even, then the last event should not be a success. So if we denote by $X_{n,p}$ a random variable with distribution $B(n,p)$, we find:
$P(X_{n,p}\text{ is even})=P(X_{n-1,p}\text{ is odd})*p+P(X_{n-1,p}\text{ is even})*(1-p)$.
Now we are ready to prove the claim, by using induction on n:
For $n=0$, the result should be clear, since then we will always have $0$ (an even number) of successes, hence the chance of $X$ being even is $1$, agreeing with the formula.
Now suppose it holds for $n-1$. Then by the previous result:
\begin{align}
P(X_{n,p}\text{ is even})
&=P(X_{n-1,p}\text{ is odd})*p+P(X_{n-1,p}\text{ is even})*(1-p)\\
&=(1-\frac{1}{2}(1+(1-2p)^{n-1}))p+\frac{1}{2}(1+(1-2p)^{n-1})(1-p)\\
&=p-\frac{1}{2}p-\frac{1}{2}p(1-2p)^{n-1}+\frac{1}{2}(1-p)+\frac{1}{2}(1-p)(1-2p)^{n-1}\\
&=p-p+\frac{1}{2}+\frac{1}{2}(1-2p)(1-2p)^{n-1}\\
&=\frac{1}{2}(1+(1-2p)^n).
\end{align}
Completing the proof.
