Will My File Be Safe?

Let's line up all of the 20 hard drives so that the first ten have the file on them and the second half don't.

If $n<10$ , then all of the files are safe no matter what.

For $n=10$ , then the only case in which the first ten files are all corrupted is if these exact ten get chosen. There are $20 \choose 10$ ways of selecting ten, and one of these will result in total corruption.

Our probability for $n=10$ is $\frac{1}{20 \choose 10}$

For $n=11$ , then there are $20 \choose 11$ ways of picking 11 out of the 20 hard drives to get corrupted. Ten of these need to be on our first ten hard drives, so the last corrupted hard drive can be any of the remaining ten, and we will still lose all of our data, so there are $10 \choose 1$ cases that will result in losing all of our data.

Our probability for $n=11$ is $\frac{10 \choose 1}{20 \choose 11}$

For $n=12$ , the same case applies. $20 \choose 12$ ways of picking our hard drives. Ten need to corrupt our first ten hard drives, so the remaining too can corrupt any of the remaining ten, so we have $10 \choose 2$ ways of picking the last two.

Our probability for $n=12$ is $\frac{10 \choose 2}{20 \choose 12}$ .

And we can see that for every value of $n \geq 10$ , the probability of all of our hard drives getting corrupted is

$\frac{10 \choose n}{20 \choose (10+n)}$ and each has a $\frac{1}{21}$ chance of occuring.

Thus our probability of all of our files getting corrupted is:

$\large \frac{1}{21} \sum_{n=0}^{10} \frac{10 \choose n}{20 \choose (10+n)}=.0909...$

So our probability of keeping our file is $1-.090909...$ , which is $\boxed{.909}$