I have recently asked this problem: What is the probability that $m$ is the largest number drawn?
I don't understand how one arrive at the questions posted by lulu and greedoid. I guess I don't understand what the problem asks for.
Could you provide a fuller explanation to these answers?
Perhaps working on a special case will simplify it for you.
Assume you have the 10 balls, numbered 1 to 10. You'll choose 3 balls at random (without replacement). What is the probability that the highest number among the balls you picked is 7.
You have $10 \choose 3$ ways of picking any 3 balls. How many possible draws if you restrict the max to 7? That is $7 \choose 3$. However, this doesn't mean, the max will be 7 (you can very well pick balls (1,2,3) all less than 7). Now, the trick is writing $P(\max=7)==P(\max \le 7)-P(\max \le 6)$
For the values we have $P(\max = 7) = \frac{{7 \choose 3} - {6 \choose 3}}{10 \choose 3}=\frac{6 \choose 2}{10 \choose 3} = \frac18$
Try special cases for one ball draw, 10 ball draw and/or max=10 to get more intuition.