Binomial Distribution vs Geometric Distribution

A binomial distribution counts successes across a fixed number of trials that are binary, independent, and identical in success probability. Write it $X \sim B(n, p)$: $n$ trials, success probability $p$ on each one, and $X$ the number of successes, which can be any whole number from 0 to $n$. The probability of exactly $k$ successes is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $\binom{n}{k}$ (n choose k) counts the orders in which those $k$ successes could fall. The mean is $\mu_X = np$ and the standard deviation is $\sigma_X = \sqrt{np(1-p)}$.

Flip a fair coin 10 times, so $n = 10$ and $p = 0.5$. Then $P(X = 6) = \binom{10}{6}(0.5)^6(0.5)^4 = \frac{210}{1024} = 0.2051$. The mean is $\mu_X = 10(0.5) = 5$ heads and the standard deviation is $\sigma_X = \sqrt{10(0.5)(0.5)} = \sqrt{2.5} \approx 1.581$ heads.

Now the error that costs the most marks: "so the probability of 6 or more heads is 0.2051." It is not. That number is the probability of exactly 6. At least 6 means 6, 7, 8, 9 or 10, and adding those five probabilities gives 0.3770, nearly double. Circle the words at least, at most, more than and fewer than before any arithmetic starts, because the formula answers only the exactly question and every other question is assembled out of it.

Independence is the condition that breaks most often in practice. Deal 10 cards from a deck without replacement and the chance of a heart shifts with every card removed, so the count of hearts is not binomial. It is close enough to treat as binomial when the sample is a small fraction of the population, which is what the 10 percent condition checks.

The binomial distribution is topic 2.10 of Unit 2, Probability, Random Variables, and Probability Distributions.

The geometric distribution describes the trial on which the first success arrives, in a run of independent trials that each succeed with the same probability $p$. Its values are 1, 2, 3 and onward with no upper limit, the probabilities are $P(X = k) = (1-p)^{k-1}p$, and the summaries are $\mu_X = \frac{1}{p}$ (mu-X, the mean) and $\sigma_X = \frac{\sqrt{1-p}}{p}$ (sigma-X).

Roll a fair die until the first 6, so $p = 1/6$. The first 6 lands on roll 3 with probability $(5/6)^2(1/6) = 25/216 \approx 0.1157$, the mean is $\mu_X = 6$ rolls, and the standard deviation is $\sigma_X = \frac{\sqrt{5/6}}{1/6} \approx 5.48$ rolls, nearly as large as the mean itself.

"The mean is 6, so the first 6 usually turns up somewhere around roll 6" is wrong three times over. The most likely single trial is the first one, $P(X = 1) = 0.1667$, and every later trial is less likely than the one before it. The median is 4, since $P(X \le 3) = 0.4213$ and $P(X \le 4) = 0.5177$. And the mean sits above the median because the right tail runs forever: by roll 6, the mean itself, only $1 - (5/6)^6 = 0.6651$ of runs have finished.

The distribution also forgets. Roll ten times with no 6 and the count of further rolls you need still has mean 6, exactly as it did at the start, because the die keeps no record. "I am due for a 6" has no arithmetic behind it.

College Board removed the geometric distribution from AP Statistics with the Fall 2026 revision, on its published list of removed topics, so it is not assessed on the exam that first runs in May 2027. It stays standard college material, and reading a waiting-time setting correctly is still how you rule out a binomial one, which is assessed at topic 2.10 of Unit 2.