In Erdos-Renyi random graph model $G(n,p)$, set $Q$ any graph property.
Suppose there exist $p_1(n)$ and $p_2(n)$ in $(0,1)$ for $n \in \mathbb{N}$ such that $Pr(G(n,p_1)\ \text{has property}\ Q) = \frac{1}{11}$ and $Pr(G(n,p_2)\ \text{has property}\ Q) = \frac{10}{11}$. Does there exist $p(n) \in (0,1)$ such that $Pr(G(n,p)\ \text{has property}\ Q) = \frac{1}{2}$?
I really have no idea how to start. One possibility is to use intermediate value theorem and then applying continuity of polynomial functions, i.e., for fixed $n$, $p \mapsto Pr(G(n,p)\ \text{has property}\ Q)$ is continuous and strictly increasing.
Now if $Q$ an increasing graph property (if $G \in Q$, then the graph obtained by adding any arbitrary edge to $G$ is also in $Q$) such that the null graph on $n$ vertices is not in $Q$ and $K_n$ is in $Q$, then $$ p \rightarrow f(p):=Pr ( G(n,p) \in Q ) $$ is a strictly increasing function with $f(0)=0$ and $f(1)=1$. This is because for fixed $n$, $$ f(p) = \sum_{G \in Q, \atop V(G) = [n]} p^{e(G)} (1-p)^{{n \choose 2}-e(G)}, $$ where $e(G)$ is the number of edges of $G$, is a polynomial in $p$ of degree at most ${n \choose 2}$. In the case that $Q$ is an increasing graph property, then $f(p)$ is non-decreasing. Non-constant, non-decreasing polynomials are strictly increasing. Since $f(p)$ is continuous, there would be a unique value $p^*$ where $f(p^*) = 1/2$.
However, your problem statement does not say that $Q$ is an increasing graph property and so $f(p)$ need not be increasing. But as before $f(p)$ is a polynomial in $p$ and hence continuous. Using the Intermediate Value Theorem as you suggested will finish off the proof.