Im trying to find the value of $p$ that will maximize $P(X=k)$ So i was thinking i sould derivative the probability mass function
$\frac{d}{dp}{n \choose k}\cdot p^{k}\cdot(1-p)^{n-k}=0$
but i cant think of any idea how to start. Im only getting to this : $k{n \choose k}p^{k-1}(1-p)^{n-k}-{n \choose k}p^{k}\left(n-k\right)\left(1-p\right)^{n-k-1}=0$
after some calculation Im getting closer to this:
$\frac{k}{n-k}=\frac{p}{1-p}$
$p=\frac{k}{n}$
Thank you for the help in the comment section..