Hypergeometric Distribution Derivation

If there are 𝑁 balls total, of which 𝐾 are successes and 𝑁-𝐾 are failures, and 𝑛 balls are randomly selected WITHOUT REPLACEMENT, then the probability that exactly 𝑘 of 𝑛 balls are success is given by:$P(X=k)=\frac{{(}_K{C}_k){(}_{N-K}{C}_{n-k})}{{}_N{C}_n}$

To see where this formula comes from, label the balls:

• 𝑆 = {𝑅₁, 𝑅₂, …, 𝑅_𝐾, 𝑊_𝐾+1, 𝑊_𝐾+2, …, 𝑊_𝑁}

Now our sample space 𝒮 consists of all subsets of 𝑆 of size 𝑛:

• 𝒮 = {𝑠⊆𝑆 | |𝑠|=𝑛}

The size of 𝒮 is equal to the number of ways to choose 𝑛 distinct balls from a size of 𝑁:

• $\mathcal{S}=\frac{N!}{n!(N-n)!}$

This is where the denominator comes from.

Next we need to enumerate those 𝑛-subsets in which exactly 𝑘 successes. To do this, we note that any such set would not only have exactly 𝑘 successes, but also exactly (𝑛-𝑘) failures. How many ways are there to choose 𝑘 successes from 𝑆 and 𝑛-𝑘 failures from 𝑆? There are:

• (_𝐾𝐶_𝑘) ways to choose 𝑘 successes
• (_𝑁-𝐾𝐶_𝑛-𝑘) ways to choose 𝑛-𝑘 failures

The choices above are independent from each other; thus the toal number of desired outcomes is simply the product of each individual event, so the total number of ways to have exactly 𝑘 successes from 𝑛 samples is:

• (_𝐾𝐶_𝑘) (_𝑁-𝐾𝐶_𝑛-𝑘)

The numerator probability