MISSION · Data and probability
The Binomial Distribution
Counting successes when every trial is the same coin
LEVEL 1 · EXPLORE
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LEVEL 2 · LEARN
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The idea in one sentence
Repeat the same two-outcome experiment \(n\) times and the binomial distribution gives the probability of getting exactly \(r\) successes.
\[P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\]The four conditions
A situation is binomial only if all four hold:
- A fixed number of trials, \(n\).
- Each trial has exactly two outcomes — success or failure.
- The probability of success, \(p\), is the same every time.
- The trials are independent.
Flipping a coin 20 times qualifies. Drawing 5 cards without replacement does not — removing a card changes the odds for the next, breaking conditions 3 and 4. Testing items from a production line qualifies, as long as the line’s failure rate is stable.
Checking these four first saves applying the formula to something it doesn’t describe.
Reading the formula
Each piece does a job.
\(p^r\) — the probability of \(r\) successes happening. \((1-p)^{n-r}\) — the probability of the other trials failing. \(\binom{n}{r}\) — the number of different orders those successes could come in.
That last factor is the part people leave out. Three heads in five flips could be HHHTT, HTHTH, TTHHH, and seven other arrangements, all equally likely. Multiplying by \(\binom{5}{3} = 10\) counts them all.
That’s also why the coefficients are the rows of Pascal’s triangle: same counting problem, same numbers.
The shape
Plot the probabilities for each \(r\) and a pattern appears.
With \(p = 0.5\) the distribution is symmetric, peaking in the middle. With \(p\) small it leans left, with most of its weight on low counts. With \(p\) large it leans right.
As \(n\) grows the shape smooths into a bell, whatever \(p\) is — which is the central limit theorem at work, and the reason the normal distribution can approximate the binomial for large \(n\).
Mean and spread
\[\text{mean} = np \qquad \sigma = \sqrt{np(1-p)}\]The mean is the obvious one: 100 flips of a fair coin average 50 heads. The standard deviation is 5, so results between 40 and 60 are unremarkable and 70 heads would be a genuine surprise — four standard deviations out.
Knowing \(\sigma\) is what turns “that seems like a lot” into a number, which is exactly what a hypothesis test needs.
“At least” questions
\(P(X \ge 1)\) over many trials is tedious directly and trivial by complement:
\[P(X \ge 1) = 1 - P(X = 0)\]One case instead of many. Reach for the complement whenever a question says “at least one”.
Common mix-ups
- Dropping \(\binom{n}{r}\), which ignores the orderings and undercounts badly.
- Using it without replacement. That’s a different distribution.
- Confusing \(r\) with \(n\) in the exponents. Successes take \(p\), the rest take \(1-p\).
- Computing “at least” the long way instead of using the complement.
Check yourself
A fair die is rolled 4 times. What’s the probability of exactly two sixes?
\(n = 4\), \(p = \frac{1}{6}\), \(r = 2\): \(\binom{4}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^2 = 6 \times \frac{25}{1296} \approx 0.116\).
LEVEL 3 · REFLECT
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