Answer
$\frac{235620}{2598960}\approx0.0907$
Work Step by Step
We know that if we want to select $r$ objects out of $n$ disregarding the order, we can do this in $_nC_r=\frac{n!}{r!(n-r)!}$ ways.
Here we have $_{52}C_5=\frac{52!}{5!47!}=2598960$ possible outcomes. Out of these $_{4}C_1\cdot_{36}C_4=235620$ satisfy the requirements (we choose $1$ ace out of the $4$ and then $4$ non ace, non picture cards). Hence the probability: $\frac{235620}{2598960}\approx0.0907$