There are $5$ vertices labelled $1,2,3,4,5$. For any two pairs of vertices $u, v$, the edge $uv$
is drawn with probability $1/2$. If the probability that the resulting graph is a tree is given by $\dfrac{p}{q}$ where $p, q$ are coprime, then find the value of $q^{1/10} + p$.
1. **Number of labeled trees on \( n \) vertices**: According to Cayley's formula, the number of labeled trees on \( n \) vertices is given by \( n^{n-2} \). For \( n = 5 \), the number of labeled trees is:
   \[
   5^{5-2} = 5^3 = 125
   \]

2. **Total number of graphs on \( n \) vertices**: The total number of graphs on \( n \) vertices, where each edge is included with probability \( \frac{1}{2} \), is \( 2^{\binom{n}{2}} \). For \( n = 5 \), the number of possible edges is:
   \[
   \binom{5}{2} = 10
   \]
   Therefore, the total number of graphs is:
   \[
   2^{10} = 1024
   \]

3. **Probability that the graph is a tree**: The probability that a randomly chosen graph is a tree is the ratio of the number of labeled trees to the total number of graphs. Thus, the probability is:
   \[
   \frac{5^3}{2^{10}} = \frac{125}{1024}
   \]

4. **Simplifying the fraction**: The fraction \(\frac{125}{1024}\) is already in its simplest form because 125 and 1024 have no common factors other than 1. Therefore, \( p = 125 \) and \( q = 1024 \).

5. **Calculating the final expression**: We need to find the value of \( q^{1/10} + p \). Given \( q = 1024 \), we have:
   \[
   1024^{1/10} = (2^{10})^{1/10} = 2
   \]
   Therefore, the final expression is:
   \[
   2 + 125 = 127
   \]

The final answer is \( \boxed{127} \).