imc

We have $1\times n=n<1+n$, hence if at least one of the numbers is $1$, the sum is larger. There $15$ such possibilities. We have $2\times 2=2+2$. For $n>2$ we already have $2\times n=n+n>2+n$, hence all other cases are good. Out of the $8\times 8$ possible cases, we find that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ cases the sum is smaller, satisfying the condition. Therefore the answer is $\frac{48}{64}=\boxed{\frac{3}{4}}$.