Find the probability of the sum of two chips equal to $10$ Ten chips numbered $1$ through $10$ are mixed in a bowl. Two chips numbered (X,Y) are drawn from the bowl, successively and without replacement. What is the probability that $X +Y = 10$?
Answer: $\frac{4}{45}$
My attempt:
The possible ways to satisfy the condition would be $(1, 9); (2, 8); (3, 7); (4, 6)$ and the same numbers with the other order. And all the possible ways to drawn two chips from the bowl would be $10\times9$. So the asked probability is given by:
$P=\frac{8}{90}=\frac{4}{45}$
I think that my attempt is reasonable, I just want to know if problems like this could be solve in a more general way, without actually having to know all the ways the condition is satisfied.
 A: 
The possible ways to satisfy the condition would be $(1, 2); (2, 8); (3, 7); (4, 6)$ and the same numbers with the other order.

Well, clearly you meant $(1, 9); (2, 8); (3, 7); (4, 6)$, but yes.
Thus $8$ among $90$ equally-probable outcomes are favoured, and so the probability is $4/45$.

I think that my attempt is reasonable, I just want to know if problems like this could be solve in a more general way, without actually having to know all the ways the condition is satisfied.

Your attempt is reasonable, and correct.   It is also the easiest method to either list the results you need to count (when the list is short enough) or otherwise have an algebraic method for counting (as you did to count the total outcome set).
A: Another way to see the problem:

*

*How many ways to sample 2 distinct chips from 10 without replacement? Ans: $10\choose 2$


*How many 2 chips samples from 10 chips with sum $10$? Ans: $4$, as you've shown.


*Therefore, $\text{Pr}(\text{sum 10})=\dfrac{4}{10\choose 2}=\dfrac{4}{45}.$
