Zoran Skoda expectation of number of valid tickets

Suppose that there are $k$ valid and $l$ invalid tickets in a pool and we choose $n$ tickets from the pool at random. What is the expectation of the number of tickets ?

The probability that a randomly chosen ticket is valid is $\frac{k}{k+l}$, so that the expectation for the number $v$ of valid tickets in the set of $n$ is $n\frac{k}{k+l}$. Indeed, this is the expectation of the sum of validity points for $n$ tickets.

We can directly find the expectation by the combinatorics. Indeed, the probability to have $i$ valid tickets in the set of $n$ is

$$P(\{v=i\}) = \frac{\binom{k}{i}\binom{l}{n-i}}{\binom{k+l}{n}},$$

so that the expectation is

$$E[v] = \frac{\sum_{i=0}^n i\binom{k}{i}\binom{l}{n-i}} {\binom{k+l}{n}} = n\frac{k}{k+l}$$

thanks to the identity

$$\sum_{i=0}^n i\binom{k}{i}\binom{l}{n-i} = \frac{n k}{k+l}\binom{k+l}{n}$$

For example, for $k=8$, $l=13$ the sum

$$\sum_{i = 0}^6 \binom{8}{i}\binom{13}{6-n}$$

can be evaluated as

$$6\cdot 28+5\cdot 56\cdot 13+4\cdot 70\cdot 78+3\cdot 56\cdot 286+2\cdot 28\cdot 715+1\cdot 8\cdot 1287+0\cdot 1716$$

what is equal to

$$124032 = \frac{16}{7} 54264 = \frac{6\cdot 8}{21}\binom{21}{6}$$

Thus, $E[v] = 16/7$. Similarly,

$$\sum_{i=0}^4 i \binom{8}{i}\binom{24}{4-i} = \binom{32}{4}$$

$$\sum_{i=0}^4 i \binom{4}{i}\binom{28}{4-i} = \frac{1}{2}\binom{32}{4}$$