Probability of taking $0$ vertex among $8$ vertices is $^{8}\textrm{C}_{0}$ [means no edge]
Probability of taking $1$ vertex among $8$ vertices is $^{8}\textrm{C}_{1}$ [means no edge]
Probability of taking $2$ vertex among $8$ vertices is $^{8}\textrm{C}_{2}$ [means $1$ edge]
.......
Now we can take any number of edges
So, number of vertices we can select=$^{8}\textrm{C}_{0}+^{8}\textrm{C}_{1}+^{8}\textrm{C}_{2}+^{8}\textrm{C}_{3}+............+^{8}\textrm{C}_{8}=2^{8}$
But according to question there is an edge with $8$ vertices , i.e. $^{8}\textrm{C}_{2}$
So, answer should be $\frac{^{8}\textrm{C}_{2}}{2^{8}}$
If they asking for Expected value, then there must be a weight or cost (like 1/2 given in GATE question). But, here that is not the scenario.
https://gateoverflow.in/1535/gate2013-24
https://www.quora.com/What-is-the-difference-between-probability-and-expectation