Numbers less than average

Question

We have a list of $n$ non-negative numbers, $\{\alpha_{1}, \alpha_{2}, \cdots \alpha_{n}\}$. It is given that $\sum_{i=1}^{n} \alpha_{i} = 1$.

If $m$ numbers are picked uniformly at random from the list, what is the probability that each of the numbers will be less than the average of the $n$ numbers in the list?

Answer 1

There is no definite answer. Consider $A = \{0.1,0.1,0.8\}$; $B = \{0.4,0.4,0.2\}$.

Then the probability for $A$ is $\dfrac{2}{3}$, while for $B$ it is $\dfrac{1}{3}$.

Answer 2

Are the numbers in the list a given or are they not but do we know how they are generated? And I assume from the question out that you take $m$ numbers without putting each of the numbers back in the list?