<s>[INST] What is the probability that two numbers between 1 and 10 picked at random sum to a number greater than 5? We have the numbers $1$ through $10$ in a box, we pick one at random, write it down and put it back in the box. We pick another of those numbers at random and write it down again. If we add the two numbers, what is the probability that it will be greater than $5$?
At first I though that I could count the number of ways we could add two numbers to get six, i.e. $2+4$ and see what are the chances to get numbers bigger than those choices. Then adding all the probabilities that relate to each way. However, I get numbers greater than $1$ which is impossible. I also thought about the chance of getting a $1$ and then a number equal to or bigger to $5$, $P(x \ge 5) = \frac 12$ multiplying them together and repeating until all numbers run out. Again, wrong answer. 
My question is: how do we get to the correct answer? Is it possible to generalize? Say that the probability of $n$ numbers picked at random from $N$ choices add to something greater than $k$.
 [/INST] You could solve this with generating functions. The generating function for this situation, equivalent to rolling a fair 10-sided die twice, is:
$$(x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^2$$
which expands to 
$$x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 5 x^6 + 6 x^7 + 7 x^8 + 8 x^9 + 
 9 x^{10} + 10 x^{11} + 9 x^{12} + 8 x^{13} + 7 x^{14} + 6 x^{15} + 5 x^{16} + 
 4 x^{17} + 3 x^{18} + 2 x^{19} + x^{20}$$
The coefficient of each $x^n$ is the number of ways of getting a sum of $n$ from the two random draws. There are 100 total possibilities ($10 \cdot 10$) with two draws from 1...10 with replacement. Looking at the polynomial above, the coefficients of the monomials $x^2$ through $x^5$ show that there are a total of 10 ways to get a sum of 5 or less. Thus, the probability of a sum greater than 5 is 90/100 or 0.9. 
Perhaps a simpler way is to first eliminate the cases where at least one draw is 5 or greater, since these guarantee a sum greater than five. There are 84 of these, leaving only the 16 cases in which both draws are 4 or less. This can also be done with a generating function (though counting manually is easy too):
$$(x + x^2 + x^3 + x^4)^2 = x^2 + 2 x^3 + 3 x^4 + 4 x^5 + 3 x^6 + 2 x^7 + x^8$$
once again giving 10 out of 100 cases where the sum is 5 or less.
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