A bag has 40 tickets number from 1 to 40. 4 tickets namely t1,t2,t3,t4 have been taken and arranged as t1<t2<t3<t4.
Then what is the probability of having t3 as 20.
What i have tried is, t1 can range from 1 to 19. so the t1's probability is 19/40.
t2 can be remaining 18 numbers which are less than 20, and total tickets are 39 now. so 18/39.
And t3 is 1/38 and t4 can be any one of the remaining 37 tickets so 1/37.
The probability of the interested event is 19/40 * 18/39 * 1/38 * 20/37.
But, this is not matching with the answer.
Please correct if my approach is wrong.