关于urn问题、不放回抽球全白球概率计算及超几何分布原理的技术问询

Great question! Let's break this down step by step to clarify both your formula validity and the core logic behind the hypergeometric distribution.

First: Is Your Formula Valid? #

Short answer: Absolutely, and we can simplify it to make its intuition clearer.

Your formula is:
$$P(X=r) = \frac{\binom{b}{r} \binom{n-b}{r-r}}{\binom{n}{r}}$$

Since $r-r=0$, and the number of ways to choose 0 non-white balls is $\binom{n-b}{0}=1$, this simplifies to:
$$P(X=r) = \frac{\binom{b}{r}}{\binom{n}{r}}$$

This makes perfect sense with classical probability principles:

• The denominator $\binom{n}{r}$ represents the total number of equally likely ways to choose $r$ balls from $n$ total balls.
• The numerator $\binom{b}{r}$ counts the number of favorable outcomes: selecting all $r$ balls from the $b$ available white balls.

This aligns perfectly with the hypergeometric distribution when we're looking for exactly $r$ "successes" (white balls) in $r$ draws.

Second: Why Multiply by Non-White Ball Combinations in the Hypergeometric Distribution? #

The hypergeometric distribution exists to model sampling without replacement from a finite population split into two distinct groups (successes/failures, white/non-white here). Its general formula is:
$$P(X=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$$
Where:

• $N$ = total population size (your $n$ balls)
• $K$ = number of successes in the population (your $b$ white balls)
• $n$ = number of draws (your $r$ balls)
• $k$ = number of successes in draws (your $r$ white balls)

The multiplication step comes down to counting all complete favorable combinations:

1. First, we choose $k$ successes from the $K$ available: that's $\binom{K}{k}$ ways.
2. Then, we need to fill the remaining $(n - k)$ draws with failures (non-white balls) from the $(N - K)$ available failures: that's $\binom{N-K}{n-k}$ ways.

These two choices are independent—picking white balls doesn't restrict how we pick non-white balls in terms of combinations—so we multiply them to get the total number of favorable mixed outcomes.

In your specific case, since you want all $r$ draws to be white, the number of non-white balls needed is $0$, so that term becomes 1. But in the general case (e.g., "probability of exactly 2 white balls in 5 draws"), we have to account for the non-white balls we're also selecting—each favorable outcome is a specific mix of white and non-white balls, and we need to count every possible valid mix.

For example: If you had 10 total balls, 4 white, and wanted exactly 2 white balls in 3 draws, you'd calculate $\binom{4}{2}$ (ways to pick 2 white) multiplied by $\binom{6}{1}$ (ways to pick 1 non-white), divided by $\binom{10}{3}$ (total ways to pick 3 balls). This gives you all 3-ball combinations that have exactly 2 white balls.

备注：内容来源于stack exchange，提问作者HellBoy