A bag contains 100 balls numbered from 1 to 100. Four balls are removed at random with out replacement. The probability that the number on the last ball is smaller than the number on first ball, is equal to

$\frac{1}{2}$

If $n_1, n_2, n_3$ and $n_4$ be the numbers on the drawn balls (in that order) than, we must have

$n_1>n_4$

Total number of order relations between

$n_1, n_2, n_3$ and $n_4$ is 4! i.e., 24

Now total number of favourable order relations have to counted.

(i) If ' $n_1$ ' is the largest, then favorable order relations are 3!

= 6

(ii) If $n_1$ is the second largest, then favourable order relations are 2(2!)

= 4

(iii) In $n_1$ is the third largest, then favourable order relations are 2!

= 2

Thus, required probability $=\frac{12}{24}=\frac{1}{2}$

Alternate solution

Clearly $n_1$ and $n_4$ are unequal. Thus there are only two cases

$n_1<n_4$  or  $n_1>n_4$

Thus, required probability $=\frac{1}{2}$