Let $E_{xy}$ be the event that $xy$ is an edge of $G$ and let $C$ be the event that $G$ is connected. By Bayes' Theorem, $$P(E_{xy} \mid C) = \frac{P(C \mid E_{xy})P(E_{xy})}{P(C)} = p\frac{P(C \mid E_{xy})}{P(C)}.$$ So the question amounts to estimating how much knowing that $G$ has an edge improves the probability of $G$ being connected.

The graph $G$ is connected precisely if it contains a spanning tree $T$. Furthermore, knowing that $xy$ is an edge of $G$, we may further require that $T$ contains the edge $xy$. So we have a crude upper bound $$P(C \mid E_{xy}) \leq \sum_{xy \in T} P(T \subseteq G \mid E_{xy}),$$ where $T$ ranges over all trees on the labeled vertex set $\{1,\ldots,n\}$. By Cayley's Formula, there are $n^{n-2}$ such $T$ and, since each such $T$ has exactly $n-1$ edges, there are $2n^{n-3}$ such trees that contain the edge $xy$. For each $T$ containing $xy$, $P(T \subseteq G \mid E_{xy}) = p^{n-2}$, so $$P(C \mid E_{xy}) \leq 2n^{n-3}p^{n-2},$$ or equivalently $$P(E_{xy} \mid C) \leq \frac{2n^{n-3}p^{n-1}}{P(C)}.$$ Clearly, this bound is only useful for small $p$. When $p$ is large, the trivial bound $$P(E_{xy} \mid C) \leq \frac{p}{P(C)}$$ is of greater use.

The classic paper
*Gilbert, E.N.*, **Random graphs**, Ann. Math. Stat. 30, 1141-1144 (1959). ZBL0168.40801, contains exact formulas for $P(C)$. Gilbert's method can also be used to compute $Q_n = P(C \mid E_{xy})$ using the recurrence $$1-Q_n = \sum_{k=2}^{n-1} \binom{n-2}{k-2}Q_k(1-p)^{k(n-k)}.$$ (The $k$th term is the conditional probability that the connected component containing $x,y$ has exactly $k$ elements.) Thus

$Q_{2} = 1$

$Q_{3} = -p^2
+ 2 p$

$Q_{4} = -2 p^5
+ 9 p^4
- 14 p^3
+ 8 p^2$

$Q_{5} = 6 p^9
- 48 p^8
+ 162 p^7
- 298 p^6
+ 318 p^5
- 189 p^4
+ 50 p^3$

I haven't tried to prove it but it seems that the bound $2n^{n-3}p^{n-2}$ from above is exactly the leading term of $Q_n$. At least this is the case for $n \leq 60$ by direct calculation.

anygiven function $p\colon \omega\to(0,1)$ ? If $p$ is any function $p\colon \omega\to(0,1)$ such that $p\gg\frac{\log n}{n}$, then $\lim_{n\to\infty}\frac{\mathbb{P}_{n,p(n)}[E_{x,y}\mid\text{$G$ is connected}]}{p(n)} = 1$. If, however, you 'are' 'inside' 'the' 'phase-transition' for the connectedness, then giving a thorough answer would require a bit of work, I think. $\endgroup$ – Peter Heinig Oct 7 '17 at 15:10youare making the question much more difficult than it is. $\endgroup$ – François G. Dorais♦ Oct 7 '17 at 20:59thoughtI had made it abundantly clear that I was imputing intentions, and that I was pointing out to the opening poster that they should not adopt my suggested additional hypotheses merely for the sake of making it a hard question, rather should make the clarifications fit whatever actual intents they have. $\endgroup$ – Peter Heinig Oct 8 '17 at 6:02