Three distinct positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?
Is there any way to find it efficiently? I'd taken example of $1,2,3,4,5,6,7,8,9$ then there are a lot of combinations to give a number divisible by three which give an integer average value. It's difficult to find the way :D
Thank you.
On
The average is an integer if the sum is divisible by $3$, which is the case if and only if the residues of the integers modulo $3$ are either all different or all the same. The probability for them to be all different is
$$ \frac{\binom31^3}{\binom93}=\frac9{28}\;, $$
and the probability for them to all be the same is
$$ \frac3{\binom93}=\frac1{28}\;. $$
Thus the probability for the average to be an integer is $\frac{1+9}{28}=\frac5{14}$.
There will be three each of integers that are $0,1,2\bmod3$ in the given range. The admissible selections sum to $0\bmod3$, of which there are four ways to achieve it: $$000\ (1\text{ selection})$$ $$111\ (1)$$ $$222\ (1)$$ $$012\ (27)$$ This gives 30 admissible selections out of $\binom93=84$ total selections, for a probability of $\frac{30}{84}=\frac5{14}$.