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
Answer & explanation
Correct answer: option 4
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 $=\frac{1}{2}$