If $12$ identical balls are to be placed in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls, is

Four numbers are chosen at random (without replacement) from the set $\left\{1,2,3,\dots \dots ,20\right\}$

Statement I: The probability that the chosen numbers when arranged in some order will form an $AP$ is $\frac{1}{85}$.

Statement II: If the four chosen numbers form an $AP$, then the set of all possible values of common difference is $\left\{\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\right\}$.