Statement 1: Our of 5 tickets consecutively numbered, three are drawn at random. The chance that the numbers on them are in A.P. is 2/15. Statement 2: Out of `2n+1` tickets consecutively numbed, three are drawn at random, the chance that the numbers on them are in A.P. is `3n//(4n^2-1)dot`

Out of (2n+1) tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in AP.

Statement-1: Out of 21 tickets with number 1 to 21, 3 tickets are drawn at random, the chance that the numbers on them are in AP is (10)/(133) . Statement-2: Out of (2n+1) tickets consecutively numbered three are drawn at ranodm, the chance that the number on them are in AP is (4n-10)/ (4n^(2)-1) .

Out of 2n tickets numbered 1,2,......,2n three are chosen at random.The probability that the numbers on them are in A.P.is

Out of 15 tokens consecutively numbered from 1 to 15, 3 tokens are drawn at random. The probability that the numbers on them in AP is

If 12 tickets numbered 0,1,2,...11 are placed in a bag,and three are drawn out,then the chance that the sum of the numbers on them is equal to 12 is -

A bag contains 20 tickets numbered 1 to 20. Two tickets are drawn at random. Find the probability that both the numbers on the ticket are prime.

From 4m+1 tickets numbered as 1,2 ,….4m+1…. Three tickets are chosen at random. Find out the probability that the numbers are in A.P. with even common difference.

There are 12 tickets numbered 1 to 12. A ticket is drawn at random. Find the probability that the number on this ticket is either a multiple of 3 or 4.

.If the numbers n-2,4n-1 and 5n+2 are in A.P. then find the value of n