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 $m > 0$ ?
 A: Let $X$ denote a discrete random variables taking on nonnegative integer values
$0, 1, 2, \ldots$ with probabilities $p_0, p_1, p_2,\ldots$. Then,
$$\begin{align}
P\{X ~ \text{even}\} &= \sum_{k \geq 0} p_{2k} = p_0 + p_2 + \cdots\\
P\{X ~ \text{odd}\} &= \sum_{k \geq 0} p_{2k+1} = p_1 + p_3 + \cdots\\
\text{Adding and subtracting, we get}\hspace{0.3in}&\\
P\{X ~ \text{even}\} +
P\{X ~ \text{odd}\} &= \sum_{k \geq 0} p_k\tag{1}\\
&= 1\\
P\{X ~ \text{even}\} -
P\{X ~ \text{odd}\} &= p_0 - p_1 + p_2 - p_3 + \cdots\\
&= \sum_{k \geq 0} (-1)^kp_k\tag{2}.
\end{align}$$
For many specific cases of random variables (including Binomial random variables),
closed-form expressions can be found for $(2)$. For example, for Poisson
random variables
$$\sum_{k \geq 0} (-1)^kp_k 
= \sum_{k \geq 0} (-1)^k e^{-\lambda}\frac{\lambda^k}{k!} 
= e^{-\lambda}\sum_{k \geq 0}\frac{(-\lambda)^k}{k!}
= e^{-2\lambda}\tag{3}$$
allowing us to conclude from $(1)$ and $(2)$ that
$$P\{X ~ \text{even}\} = \frac{1+e^{-2\lambda}}{2} = e^{-\lambda}\cosh(\lambda).$$
I will leave the similar calculation for binomial random variables to you as an exercise: the final result works out to be what Uncountable showed.
A: 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.
A: 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.
