@Arjun Sir please verify.

The largest and the smallest numbers will only be compared if either of them are chosen as the pivot in the very first split. In the subsequent splits they will never get compared since the first pivot divides the list into two and the largest and smallest number will be put in two different halves. The probability would thus be the probability to choose the largest or the smallest as the pivot in the very first step. Thus the required probability would be ${1 \over n}+ {1 \over n} = {2 \over n}$.