In a box there are $M_1$ balls numbered 1, $M_2$ numbered 2... $M_N$.

Question

In a box there are $M_1$ balls numbered 1, $M_2$ numbered 2... $M_N$. From the box $n$ balls are drawn without returns. Find the mathematical expectation of the number of numbers that are not drawn.

The answer is given, but it seems very simplified, I don;t know how correct it is. If someone understands this, I would love an explanation:

$$\sum_{i=1}^{N}{\binom{M_1+M_2+...+M_N-M_i}{n} \over \binom{M_1+M_2+...M_N}{n}}$$

Answer 1

For $i=1,\dots,N$ let $X_{i}$ take value $1$ if number $i$ is not drawn and value $0$ otherwise.

Then $X:=X_{1}+\cdots+X_{N}$ equals the number of numbers that are not drawn.

Then $\mathbb{E}X=\mathbb{E}\left(X_{1}+\cdots+X_{N}\right)=\mathbb{E}X_{1}+\cdots+\mathbb{E}X_{N}=\mathbb{P}\left(X_{1}=1\right)+\cdots+\mathbb{P}\left(X_{N}=1\right)$

An expression for $\mathbb{P}\left(X_{i}=1\right)$ is given in your question.

Answer 2

The $i$th term in the sum is the probability of drawing $n$ balls, none of which are labeled $i$. Since not drawing a given number just adds $1$ to the amount of numbers not drawn, we simply add the probability of each number not being drawn to get the expected value.