A basketball player makes 70% of her free throws. If she shoots 10 free throws in a game, what’s the probability she makes exactly 7? At least 8? This is not a question you can answer by intuition, but it’s exactly the kind of question the binomial distribution was built to answer — and once you understand its logic, dozens of AP Statistics, A-Level, and undergraduate problems that look intimidating become completely mechanical.
What Is a Binomial Distribution?
The binomial distribution describes the number of “successes” in a fixed number of independent trials, where each trial has only two possible outcomes and the same probability of success every time. It is a discrete probability distribution — the random variable it describes can only take whole-number values (0, 1, 2, 3, and so on), never fractions or decimals.
The word “binomial” literally means “two names,” referring to the two possible outcomes of each trial: success or failure. It doesn’t matter what “success” actually represents in context — making a free throw, a coin landing heads, a customer clicking a link, a product passing inspection. As long as the underlying structure fits, the same mathematics applies.
The Four Conditions for a Binomial Setting
Before reaching for the binomial formula, check that your situation genuinely satisfies four conditions. Students lose easy exam marks by applying the binomial formula to situations that don’t actually qualify.
1. Fixed number of trials (n). You must decide in advance how many trials will occur — 10 free throws, 20 coin flips, 50 items inspected. The number cannot depend on the outcomes themselves.
2. Two possible outcomes per trial. Each trial results in either a “success” or a “failure” — there is no third option. A quality inspection is either pass or fail; a coin is either heads or tails.
3. Constant probability of success (p). The probability of success must be the same on every trial. This is where sampling without replacement from a small population can quietly break the binomial assumption — if you draw cards from a deck without putting them back, the probability changes trial to trial, and the binomial model no longer strictly applies (a related but different distribution, the hypergeometric, is designed for exactly this case).
4. Independence. The outcome of one trial must not affect the outcome of another. Free throws are usually treated as independent (each shot doesn’t change the probability of the next), but this assumption can fail in practice — a player’s confidence after a make or miss could, in reality, shift the true probability slightly. Exam questions will typically tell you to assume independence explicitly.
The Binomial Probability Formula
If X is the number of successes in n independent trials, each with probability of success p, then X follows a binomial distribution, often written X ~ B(n, p). The probability of getting exactly k successes is:
P(X = k) = C(n, k) × pk × (1 − p)n−k
Each piece of this formula has a clear job:
C(n, k), read as “n choose k,” counts the number of different orders in which k successes could occur among n trials. It is calculated as n! / [k!(n−k)!], where the exclamation mark denotes factorial (5! = 5 × 4 × 3 × 2 × 1). This term exists because a success on trial 1 and a failure on trial 2 is a different sequence from a failure on trial 1 and a success on trial 2, even though both give exactly one success out of two trials — and the formula needs to count all such arrangements.
pk is the probability of getting k successes in a row, and (1−p)n−k is the probability of getting the remaining n−k trials as failures. Multiplying these together with the count of arrangements gives the total probability of exactly k successes, however they happen to be ordered.
Worked Example: Free Throws
Back to the basketball player who makes 70% of her free throws (p = 0.7), shooting n = 10 times. What is the probability she makes exactly 7?
P(X = 7) = C(10, 7) × (0.7)7 × (0.3)3
First, C(10, 7) = 10! / (7! × 3!) = 120. Next, (0.7)7 ≈ 0.0824, and (0.3)3 = 0.027. Multiplying all three together:
P(X = 7) = 120 × 0.0824 × 0.027 ≈ 0.267
So there’s roughly a 26.7% chance she makes exactly 7 out of 10. Now suppose the question asks for “at least 8” instead. This is a different calculation — you cannot just find P(X = 8) and stop, because “at least 8” means 8, 9, or 10:
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
Calculating each term the same way: P(X = 8) ≈ 0.2335, P(X = 9) ≈ 0.1211, and P(X = 10) ≈ 0.0282. Summing these gives P(X ≥ 8) ≈ 0.383, or about a 38.3% chance. Notice how different this is from P(X = 8) alone — confusing “exactly” with “at least” or “at most” is one of the most common errors on binomial problems, covered again below.
Mean and Variance of a Binomial Distribution
Rather than summing the full probability formula every time, two simple shortcuts give you the centre and spread of a binomial distribution directly:
Mean: μ = np Variance: σ² = np(1 − p) Standard deviation: σ = √[np(1 − p)]
These formulas make intuitive sense. If you shoot 10 free throws with a 70% success rate, you’d expect to make 10 × 0.7 = 7 on average — exactly matching np. The variance formula captures something less obvious: spread is maximised when p = 0.5 (maximum uncertainty about each trial) and shrinks toward zero as p approaches 0 or 1 (near-certain outcomes on every trial).
Worked Example: Mean and Variance
For the free-throw shooter (n = 10, p = 0.7):
μ = 10 × 0.7 = 7 makes (on average, across many 10-shot sessions)
σ² = 10 × 0.7 × 0.3 = 2.1
σ = √2.1 ≈ 1.45 makes
So while 7 is the expected number of makes, a standard deviation of about 1.45 tells you that outcomes anywhere from roughly 5 to 9 makes would be unsurprising in a typical 10-shot session — the earlier calculation showing a combined 38.3% chance of 8 or more makes fits neatly with this picture of spread around the mean.
When the Binomial Distribution Approximates the Normal Distribution
As n gets large, the binomial distribution starts to look increasingly like a smooth, symmetric bell curve — this is a direct consequence of the Central Limit Theorem. A commonly used rule of thumb is that the normal approximation is reasonable when both np ≥ 10 and n(1−p) ≥ 10. For the free-throw example, np = 7 and n(1−p) = 3, both below 10, so a normal approximation would not be reliable here and the exact binomial calculation above is the right approach.
When the approximation is valid, you can estimate binomial probabilities using a normal distribution with mean np and standard deviation √[np(1−p)] — which is far faster than summing binomial terms by hand for large n, and is the technique many statistical software packages use internally.
Binomial vs Related Distributions
| Distribution | Use When… |
|---|---|
| Binomial | Fixed number of independent trials, constant p, counting successes |
| Poisson | Counting rare events over a continuous interval of time or space, no fixed “n” |
| Hypergeometric | Sampling without replacement from a finite population — probability changes each draw |
| Geometric | Counting trials until the first success, rather than successes in a fixed n |
Real-World Applications
The binomial distribution appears constantly outside the classroom. In quality control, a factory testing a batch of 200 items for defects, where each item independently has a 2% defect probability, uses the binomial model to set acceptable thresholds for how many defects should trigger a line shutdown. In clinical trials, researchers use binomial reasoning to determine how many patients out of a treatment group would need to respond positively before a drug is considered effective, given some baseline response rate. In marketing analytics, an email campaign sent to 5,000 people with a historical 3% click-through rate can be modelled as binomial to estimate the probability of getting at least a certain number of clicks — useful for setting realistic campaign targets before the results come in.
Common Mistakes
Confusing “exactly,” “at least,” and “at most.” As shown above, these require entirely different calculations. “At least k” means summing P(X=k) through P(X=n); “at most k” means summing P(X=0) through P(X=k); “exactly k” is the single term. Always read the question wording twice before setting up the sum.
Forgetting the combination term. Students sometimes calculate pk(1−p)n−k and stop, forgetting to multiply by C(n,k) — this gives the probability of one specific ordering of successes and failures, not the total probability of getting k successes in any order.
Applying the binomial model when trials aren’t independent or p isn’t constant. Sampling without replacement from a small population, or situations where earlier outcomes plausibly change later probabilities, technically violate the binomial assumptions — even though the binomial is often used as a reasonable approximation in these cases.
Practice Question
A multiple-choice test has 6 questions, each with 4 answer options. A student guesses randomly on every question. What is the probability the student gets exactly 2 questions right?
Answer: This is binomial with n = 6, p = 0.25 (one correct option out of four), k = 2.
P(X=2) = C(6,2) × (0.25)² × (0.75)&sup4;
C(6,2) = 15. (0.25)² = 0.0625. (0.75)&sup4; ≈ 0.3164.
P(X=2) = 15 × 0.0625 × 0.3164 ≈ 0.297, or about a 29.7% chance of getting exactly 2 out of 6 right by pure guessing.
References
1. Moore, D.S., McCabe, G.P. and Craig, B.A. (2021) Introduction to the Practice of Statistics. W.H. Freeman.
2. Triola, M.F. (2022) Elementary Statistics. Pearson.
3. Ross, S.M. (2020) A First Course in Probability. Pearson.
4. Wackerly, D., Mendenhall, W. and Scheaffer, R.L. (2014) Mathematical Statistics with Applications. Cengage.
5. Field, A. (2018) Discovering Statistics Using IBM SPSS Statistics. SAGE.