Expected value of objects with a serial number greater than a given number I have $n$ objects, numbered $1$ to $n$. If I take out $m$ objects randomly, then what is the expected value of the number of objects whose serial number is greater than to $x$? $ (1 \le x \le n) $
EDIT 1: Here's my best attempt. If I denote the probability that exactly $i$ objects have a serial number greater than $x$ as $p(i)$, then the expected value is $\sum_{i=1}^{n} ip(i)$
There are $x$ objects whose serial number is not greater than $x$ and $n-x$ objects whose is. Therefore the probability that exactly $i$ objects have a serial number greater than $x$ is.... I'm not sure. I'm guessing $$ \frac{\binom{x}{m-i} \binom{n-x}{i}}{\binom{n}{m}} $$
I don't know if this is correct or whether this sum has a closed form
 A: Number the objects that are taken out.
For $i=1,\dots,m$ let $X_i$ be a random variable taking value $1$ if the object with label $i$ has serial number greater than $x$ and let it take value $0$ otherwise.
If $X:=X_1+\cdots+X_m$ then $X$ is the number of objects that has a serial number greater than $x$.
So now calculate its expectation and for that use linearity of expectation and symmetry.
The answer will appear to be surprisingly simple.
A: From your answer
The probability that exactly $i$ objects have higher number than $x$ is given by the following expression
$$
P(i)=\frac{\binom{x}{m-i}\binom{n-x}{i}}{\binom{n}{m}}
$$
Extend to obtain expected value
Now we just need to multiply each probability with $i$ and sum for all possible $i$ to obtain expected value
$$
\begin{align}
E&=\sum_{i=0}^{m} i\cdot\frac{\binom{x}{m-i}\binom{n-x}{i}}{\binom{n}{m}}
\\\\
&=\frac{(n-x)\binom{n-1}{m-1}}{\binom{n}{m}}
\\\\
&=(n-x)\frac{m}{n}
\\\\
&=
m-x\cdot\frac{m}{n}
\end{align}
$$
Some explanations
Say that you select some object with higher number than $x$, say $i$ of them, then it means you select $m-i$ objects with number less than or equal to $x$. Imagine that in addition to this, you also want to choose one of the selected objects with high number and label it special edition, the number of ways to do this is then given by
$$
\sum_{i=0}^{m}i\cdot\binom{x}{m-i}\binom{n-x}{i}
$$
Which is no other than the numerator in our summand. Another way to do this is to select the special edition object first, then select the rest of $m-1$ objects from remaining $n-1$ objects. The number of ways to do this is
$$
(n-x)\binom{n-1}{m-1}
$$
Two expressions used to count the same number of possibilities, so they must be equal
$$
\sum_{i=0}^{m}i\cdot\binom{x}{m-i}\binom{n-x}{i}
=
(n-x)\binom{n-1}{m-1}
$$
