Let $X$ be the number of connected components in $G(n,p)$. If we fix $n$ and vary $p$, is $E(X)$ a decreasing function of $p$? I "feel" that this should be right because as $p$ increases there are likely to be more edges and that has a tendency to reduce the number of components.
But is it actually true? And if so how should one prove it?
Yes, by coupling of the family $(G(n,p))_{0\leqslant p\leqslant 1}$ for every fixed $n$.
More explicitely, for each possible edge $e$, decide that $e$ is an edge of $G(n,p)$ if and only if $U_{xy}\leqslant p$, where $(U_{xy})_{xy}$ is an i.i.d. collection of random variables uniform on $(0,1)$.
Then, for $p\leqslant q$, every edge of $G(n,p)$ is an edge of $G(n,q)$ as well, hence the respective numbers $X_p$ and $X_q$ of connected components of $G(n,p)$ and $G(n,q)$ are such that $X_p\geqslant X_q$ almost surely.
Finally, if $p\lt q$, then $X_p\gt X_q$ with positive probability, hence the function $p\mapsto E[X_p]$ is strictly decreasing.