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?

Question

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.

Answer 1

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.

Answer 2

There are ${11\choose 3}=165$ ways to choose $3$ tickets. If the middle one has number $m\in[2..6]$ there are $m-1$ favorable triples, and for $m'\in[7..10]$ we obtain the same as for $m\in[2..5]$. It follows that there are $$2(1+2+3+4)+5=25$$ favorable triples. The requested probability then is $p={25\over165}={5\over33}$.