Problem:
A box contains 20 tickets numbered 1 to 20. If three tickets are chosen at random find the probability that the numbers on the drawn tickets are in arithmetic progression.
My solution:
I was unable to apply any conventional combinatorics formulae so I just applied basic arithmetical logic.
Lets assume the tickets picked are numbered x, y and z and the common difference between these is 'd' (assuming they are in A.P.). Now d cannot be greater than 9 because lets say if d is 10 then minimum x is 1 , y= x+d=1+10=11 but then z becomes z=y+d=11+10; z=21, but card numbered 21 is not present. So common difference d can only go upto 9.
Now for d=1 we can choose x, y, z as 1,2,3 or 2,3,4 or … uptill 18,19,20 hence we get 18 such sets as the first number ranges from 1 to 18. Similarly for d=2 we can chose 1,3,5 or 2,4,6 or … uptill 16,18,20 here the first number goes only uptill 16 as choosing 17 gives us 17,19,21; so for d=2 we get 16 such sets.
So the pattern we are getting is: For d=1 : 18 sets, d=2 : 16 sets …d=9: 2 sets. So now if we add up the number of sets for each common difference d we will get the total number of favourable cases i.e 18+16+…+2 = 90. Now we can find the total number of cases by 20C3 (the total number of possible ways 3 cards can be picked from 20 cards).
So final answer 90/20C3 = 0.079.
Is it wrong? If so, where is the fallacy in my logic? Also please provide the correct solution if you can.
Thank you and kind regards.
Best Answer
I have the same answer as you. But I used a different method.
The first and the third term of an arithmetic sequence must be of the same parity. Once the first and the third term are chosen, the second term must be the arithmetic mean of the other two terms. So we are choosing either two odd or two even numbers as the first and the third term. The number of combinations is
$$\binom{10}{2}+\binom{10}{2} =90$$