I kinda tried it and got the answer 5/33.
But I feel like I am doing something wrong.
What I did was:.
Firstly I found favorable outcomes. My favourable outcomes were 25 such triplets. I think I am doing something wrong here only.
Then I found total outcomes by C(11,3).
Then I found the probability.
Please anyone correct me if I am wrong.
[Math] Out of 11 tickets marked with nos. 1 to 11, 3 tickets are drawn at random. Find the probability that the numbers on them are in AP
arithmetic-progressionselementary-probabilityprobability
Best Answer
I expect the hard part is making sure you didn't miss some of the progressions. Let's count them by period.
Period $1$: $(1,2,3), (2,3,4), \cdots, (9,10,11)\,$ so $\boxed 9$
Period $2$: $(1,3,5), (2,4,6), \cdots , (7,9,11)\,$ so $\boxed 7$
Period $3$: $(1,4,7), (2,5,8), \cdots, (5,8,11)\,$ so $\boxed 5$
Period $4$: $(1,5,9), (2,6,10), (3, 7,11)\,$ so $\boxed 3$
Period $5$: $(1,6,11)\,$ so $\boxed 1$
Thus there are $1+3+5+7+9=25$ three term progressions. As there are $\binom {11}3=165$ ways to choose three with no restriction, the answer is $$\frac {25}{165}=\boxed {\frac 5{33}}$$
Confirming your result.