无放回抽样下样本个体的无条件抽样概率验证与推广问询

Hey everyone, I recently worked through a problem about sampling without replacement and managed to generalize the result, so I wanted to share this here to get some thoughts or confirm my reasoning.

Suppose that 40% of 5000 voters favor candidate Jones. A random sample of $n=10$ voters will be selected. For the first person selected, the probability of favoring Jones is 0.4. In Exercise 3.35 you will show that unconditionally the probability that the second person favors Jones is also .4. (Chapter 3, Mathematical Statistics,Wackerly)

I extended this to a more general case: Let’s say we have $N$ total voters, with $m$ voters favoring Jones and $f$ voters who don’t. To find the unconditional probability that the second selected voter favors Jones, we can sum the conditional probabilities over all possible first-selection events:

$$\frac{m}{N}\cdot \frac{m-1}{N-1}+\frac{f}{N}\cdot \frac{m}{N-1}=\frac{m(m-1+f)}{N(N-1)}=\frac{m(N-1)}{N(N-1)}=\frac{m}{N}$$

(We use the relation $m+f=N$ to simplify the expression here.)

备注：内容来源于stack exchange，提问作者Starlight