out of 11 tickets marked with no 1 to 11 three tickets are drawn at random .find probability that numbers are in AP
When draw three tickets out of 11, total possible outcomes = C(11,3) =
Total number of possible cases when :
Common difference is 1
1,2,3
2,3,4
3,4,5
5,6,7
6,7,8
7,8,9
8,9,10
9,10,11
With common difference 2
1,3,5
2,4,6
3,5,7
4.6.8
5,7,9
6,8,10
7,9,11
With common difference as 3
1,4,7
2,5,8
3,6,9
4,7,10
5,8,11
With common difference 4
1,5,9
2,4,10
3,7,11
With common difference 5
1,6,11
With common difference 6
No such APs
Total number of favourable outcomes = 25
Therefore, required probability=
Total number of possible cases when :
Common difference is 1
1,2,3
2,3,4
3,4,5
5,6,7
6,7,8
7,8,9
8,9,10
9,10,11
With common difference 2
1,3,5
2,4,6
3,5,7
4.6.8
5,7,9
6,8,10
7,9,11
With common difference as 3
1,4,7
2,5,8
3,6,9
4,7,10
5,8,11
With common difference 4
1,5,9
2,4,10
3,7,11
With common difference 5
1,6,11
With common difference 6
No such APs
Total number of favourable outcomes = 25
Therefore, required probability=