Most probable value of $x$ I was asked :


  We performed $n$ different trials in which an event $A$ can occur with probability $p$. Let $X$ be the number of occurrences of $A$. Determine the most probable value of $X$.


My try - 

$P_X(x)=\binom{n}{x}p^x(1-p)^{n-x}$

By common sense, we know that the middle term $\lfloor \frac n2 \rfloor$ much be the one with largest $nC_x$. Apart from that, I am unable to proceed further to prove that a particular $x$ is most probable.

Or am I reading the question wrong? Are they asking the Expectation? 

Thanks in advance.
Note that: $$\frac{P_X(x+1)}{P_X(x)}=\frac{p(n-x)}{(1-p)(x+1)}$$

Thus:


$P_X(x+1)&gt;P_X(x)$ when $\frac{p(n-x)}{(1-p)(x+1)}&gt;1$ or simplified:  $x&lt;pn-\left(1-p\right)$
$P_X(x+1)=P_X(x)$ when $x=pn-\left(1-p\right)$
$P_X(x+1)&lt;P_X(x)$ when $x&gt;pn-\left(1-p\right)$


$P_X$ increases up to a point and then decreases. Find the last $\tilde{x}$ for which $\tilde{x}\leq pn-\left(1-p\right)$. The next value must be (weakly) larger, but none can be larger than that. $\tilde{x}+1$ is a mode. But note there are two modes when this holds with equality. 

Some pictures:
Because we have a discrete number of events with two possible outcomes, this is just a binomial distribution. The mean number of successes $X$, given $n$ trials with probability of success $p$, is $X = np$. (This can be proved by summing over all possible numbers of successes from $0$ to $n$; see the proof on Wikipedia here.)