Discrete Counting Distributions: Binomial, Negative Binomial, Hypergeometric

A crate of 10 parts holds 4 defective. Sample 3. The chance of at most one defective is 2/3 without replacement, but 0.648 if you wrongly treat it as binomial. One key word, "without replacement," picks the distribution.

$n$ independent trials, each a success with probability $p$; $X$ counts the successes.

$$P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}, \quad k = 0, 1, \ldots, n$$

$E[X] = np$, $\text{Var}(X) = np(1-p)$. Use it when the number of trials is fixed and $p$ stays constant, which means sampling with replacement or from an effectively infinite population.

Draw $n$ items without replacement from a population of $N$ containing $K$ successes; $X$ counts the successes drawn.

$$P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$

The mean is $E[X] = nK/N$, the same as a binomial with $p = K/N$. The variance adds a finite-population correction:

Common mistakes

• Treating without-replacement sampling as binomial. The crate problem is hypergeometric (0.667), not binomial (0.648). The "without replacement" wording is the tell. Trap: 0.648.
• Mixing the negative binomial conventions. $E[X] = r(1-p)/p$ counts FAILURES before the $r$-th success. If a problem counts total TRIALS, the mean is $r/p$. Read which one is asked.
• Dropping the finite-population correction. The hypergeometric variance includes $(N-n)/(N-1)$. Using $np(1-p)$ with $p = K/N$ overstates the variance.

Bottom line

• Binomial($n, p$): successes in $n$ independent trials at a constant $p$. $E = np$, $\text{Var} = np(1-p)$; requires constant $p$ and independence.
• Hypergeometric: successes drawing $n$ WITHOUT replacement from $N$ items containing $K$. $E = nK/N$; variance carries the correction $(N-n)/(N-1)$.
• Negative binomial($r, p$): failures before the $r$-th success. $E = r(1-p)/p$, $\text{Var} = r(1-p)/p^2$; $r=1$ is geometric.
• Key-word test: with replacement (or huge population) is binomial; without replacement from a finite population is hypergeometric; "until the $r$-th success" is negative binomial.

Exam shortcut

Read for the sampling scheme before you compute. "With replacement is binomial, without replacement is hypergeometric, until the $r$-th success is negative binomial." For a negative binomial given its moments, "variance over mean is one over $p$" backs out $p$ in a single step, then the mean gives $r$.

The full lesson (about 2,113 words, 14 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 2a
• 2b
• 2c
• 2d
• 2e
• 2f

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