Answer
See below.
Work Step by Step
We know that probability$=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}.$ The number of good outcomes is $1$.
If we want to choose $k$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $_{n}C_k=\frac{n!}{(n-k)!k!}$ ways.
The order doesn't matter here when choosing the balls, thus we have to use combinations. We have $59$ balls for $5$ spots and $35$ balls for $1$ spot, thus the number of good outcomes is: $_{59}C_{5}\cdot_{35}C_{1}=\frac{59!}{(59-5)!5!}\frac{35!}{(35-1)!1!}=175223510
$
Thus probability=$\frac{1}{175223510}$
