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 $k$.
Best Answer
In this case, it is easier to find the probability $p$ that the sum is $\le 5$. Then the answer to the original problem is $1-p$.
We can list and count the ordered pairs $(a,b)$ of numbers that have sum $\le 5$.
If $a=1$ there are $4$, if $a=2$ there are $3$, and so on down to $1$.
So there are $10$ ordered pairs. Each has probability $\frac{1}{10^2}$, so $p=\frac{10}{10^2}$.
Remark: The same idea will work in your general case, as long as $k\le n+1$. A modification will take care of the rest of the $k$.