The binomial distribution shows up the moment a problem asks for the probability of a certain number of "successes" in a fixed number of independent trials. Coin flips, free throws, defective parts on an assembly line — all binomial, once you check the conditions. This walkthrough is the conditions, the formula, the mean and variance, and a full worked example.
The Four Conditions for a Binomial Setting
Before you reach for the formula, check that the situation actually fits the model. A binomial distribution applies when BINS holds:
- B — Binary outcomes. Each trial has exactly two possible results, labeled "success" and "failure."
- I — Independent trials. The outcome of any one trial does not affect any other.
- N — Fixed number of trials, n. You decide n in advance; it does not depend on the data.
- S — Same probability of success, p. The probability of success stays the same on every trial.
If any of these fails, binomial is the wrong model. A common failure is independence: drawing cards without replacement changes p as you go, so the count of Hearts in 5 draws from a 52-card deck is not binomial — it is hypergeometric. With replacement, it is binomial.
The Binomial Formula
If X is the number of successes in n independent trials, each with probability p of success, then X follows a binomial distribution with parameters n and p, written X ~ Binomial(n, p). The probability that X equals a specific value k is
P(X = k) = C(n, k) · p^k · (1 − p)^(n − k)
where C(n, k) = n! / (k!(n − k)!) is "n choose k," the number of ways to arrange k successes among n trials.
The three pieces have clean intuitions. p^k is the probability of one specific arrangement of k successes; (1 − p)^(n − k) is the probability that the remaining n − k trials are all failures; C(n, k) counts how many different orderings give exactly k successes.
The mean and variance of a binomial are short:
- E(X) = np
- Var(X) = np(1 − p)
- σ = √(np(1 − p))
These come from treating X as the sum of n independent Bernoulli(p) trials, each with mean p and variance p(1 − p).
The Worked Example
A free-throw shooter makes 70% of her attempts. She takes 10 shots. The shots are independent, p = 0.7 stays the same, n = 10 is fixed, and each shot is made or missed — all four conditions hold, so X = number of made free throws follows Binomial(10, 0.7).
Mean and variance.
- E(X) = np = 10 × 0.7 = 7
- Var(X) = np(1 − p) = 10 × 0.7 × 0.3 = 2.1
- σ = √2.1 ≈ 1.45
On average she makes 7 out of 10, with most outcomes in the 5–9 range.
P(X = 8): exactly 8 made.
P(X = 8) = C(10, 8) · (0.7)^8 · (0.3)^2 = 45 · 0.05765 · 0.09 ≈ 0.2335
C(10, 8) = 45 (choose which two shots she misses). 0.7^8 ≈ 0.05765 and 0.3^2 = 0.09. So there is about a 23% chance she makes exactly 8 of 10.
P(X ≥ 8): at least 8 made.
For "at least," sum P(X = 8) + P(X = 9) + P(X = 10).
- P(X = 9) = C(10, 9) · 0.7^9 · 0.3^1 = 10 · 0.04035 · 0.3 ≈ 0.1211
- P(X = 10) = C(10, 10) · 0.7^10 · 0.3^0 = 1 · 0.02825 · 1 ≈ 0.0282
P(X ≥ 8) ≈ 0.2335 + 0.1211 + 0.0282 ≈ 0.3828
So about 38% of the time she makes 8 or more of 10. About 62% of the time she makes 7 or fewer.
P(X ≤ 4): the complement trick.
For "at most 4," computing P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) by hand is tedious. A shorter route uses the complement when the cumulative probability is close to 1 from the upper side: P(X ≤ 4) = 1 − P(X ≥ 5). Both work; pick whichever needs the fewer terms.
When the Binomial Fails (and What to Use Instead)
Two relatives of the binomial cover the situations where one of the BINS conditions breaks:
- Hypergeometric distribution — counts of successes in n draws without replacement from a finite population (e.g., five cards from a 52-card deck). p shifts after each draw, so independence fails. The hypergeometric formula uses combinations of the population, not n^k powers.
- Negative binomial / geometric — instead of a fixed n, you keep going until a fixed number of successes (or one success, for geometric). Here it is the trial count that is random, not the success count.
If the conditions barely hold — say, p drifts slightly over trials — the binomial is often still a good approximation. If they break badly, switch models or simulate.
The Normal Approximation
For large n, the binomial distribution is approximately normal with the same mean np and standard deviation √(np(1 − p)). A common rule of thumb is that the approximation is reliable when both np ≥ 10 and n(1 − p) ≥ 10.
For the free-throw example, np = 7 and n(1 − p) = 3 — neither comfortably above 10, so the normal approximation is shaky here. With n = 100 and p = 0.7 (np = 70, n(1 − p) = 30) you could safely standardize and use a z-table to find binomial probabilities to a few decimal places. Use the exact formula when n is small and the normal approximation when n is large.
Getting Help
Once you are comfortable with binomial, the natural next step is the Poisson distribution explained, which handles counts of rare events when n is large and p is small. To build the probability rules that the binomial relies on under the hood, probability rules: addition and multiplication covers independence and the multiplication rule.
Conclusion
A binomial distribution counts successes in a fixed number of independent trials with a constant success probability. Check BINS first; if it holds, use P(X = k) = C(n, k) p^k (1 − p)^(n − k), and remember E(X) = np and Var(X) = np(1 − p). The free-throw walkthrough above produced P(exactly 8) ≈ 0.234 and P(at least 8) ≈ 0.383 — both straightforward once the formula is set up. For "at most" or "at least" problems, the complement is often the cleaner path.
