An urn with 32 balls where $5\choose r$ are labelled with the number $r$ (for $r=0,1,2,3,4,5$).

Suppose we choose randomly and without replacement 3 balls from the urn, and let $S$ be the sum of the numbers written on these 3 chosen balls then what what is the expectation value of $S$, i.e. $E(S)$?

Do I actually need to write every possible option for the sum $S$ and then multiply by a suitable probability?. For example for $S=0+1+2=3$ we have the probability is $(1/32)\times 5\times (1/31)\times 10 \times (1/30)$, and then do the same for the other options of numbers.

Seems like a lot of work, is there a shortcut I don't see? Thanks.

Solution 1:

If you draw $1$ ball, the expected value is $2.5$ (as you can check). Expectation is linear. Which means the sum of three draws (with or without replacement) is simply $3\cdot (2.5)=7.5$

Solution 2:

There is a symmetry to the problem: whenever there is a possibility $(a, b, c)$ for the values of the three draws, there is another possibility $(5-a, 5-b, 5-c)$ which has equal probability. Furthermore, if the value of the random variable on $S(a,b,c) = x$, then the value on $S(5-a, 5-b, 5-c) = 15-(a+b+c) = 15-x$.

Now, using this symmetry, we can conclude that $E(S) = 15 - E(S)$.