Difficulty: Intermediate | Prerequisites: random variables, PMF, expected value, and variance (first half of Ch. 5).
The binomial distribution is the first named discrete distribution most statistics courses cover, and it shows up relentlessly on exams. It models the number of successes in a fixed number of independent, identical trials, each with only two outcomes. If you understand the four conditions that make an experiment binomial, and you can plug into the formula, you can handle a wide range of probability questions, from quality control to medical trials. The mean and variance of a binomial have clean shortcut formulas that save considerable time.
A binomial experiment counts successes in n independent trials where each trial has the same probability p of success. The probability of exactly x successes is given by the binomial PMF. The mean is np and the variance is np(1 – p).
Binomial experiment
An experiment that satisfies four conditions (often remembered as BINS): Binary outcomes, Independent trials, fixed Number of trials, and constant probability of Success. Think of it as any scenario where you repeat the same yes/no test a set number of times under identical conditions.
Binary outcome
Each trial has exactly two possible results: success or failure. "Success" is just the outcome you are counting; it does not have to be a good thing (e.g., "defective" can be the success if you are counting defective items).
Independence (of trials)
The result of one trial does not affect the probability of success on any other trial. Drawing with replacement satisfies this; drawing without replacement from a very large population approximately satisfies it.
n (number of trials)
The fixed, predetermined count of how many times the experiment is repeated. This is set before the experiment begins.
p (probability of success)
The probability of success on any single trial. It must remain constant across all n trials.
Combination (n choose x)
The number of ways to choose x items from n, written as C(n, x) or "n choose x." Calculated as n! / [x!(n – x)!]. In simple terms, it counts the number of different orderings in which x successes and (n – x) failures could occur.
Binomial probability P(X = x)
The probability of getting exactly x successes in n trials: C(n, x) × pˣ × (1 – p)ⁿ⁻ˣ.
Before using the binomial formula, verify all four conditions hold:
B (Binary): each trial has exactly two outcomes (success/failure).
I (Independent): trials do not influence one another.
N (Number): the number of trials n is fixed in advance.
S (Success probability): p stays the same from trial to trial.
If any condition fails, the binomial model does not apply.
Worked example from the notes: a batch of 15 items where each has a 20% chance of failing a binding strength test.
Binary: each item either fails (success) or does not (failure). Yes.
Independent: one item's result does not affect another's. Yes.
n = 15, fixed. Yes.
p = 0.2, constant. Yes.
Conclusion: this is a binomial experiment.
Counter-example: a drug trial where different patients receive the drug or a placebo, and efficacy and side effects are tracked for both groups. The outcomes per patient are not binary in the simple success/failure sense, and the groups have different treatments, so the constant-p condition fails. This is not a single binomial experiment.
P(X = x) = C(n, x) × pˣ × (1 – p)ⁿ⁻ˣ, where x = 0, 1, 2, …, n.
C(n, x) = n! / [x!(n – x)!].
P(X = 0): P(X = 0) = C(15, 0) × (0.2)⁰ × (0.8)¹⁵ = 1 × 1 × (0.8)¹⁵. This is the probability that none of the 15 items fail.
P(X = 15): P(X = 15) = C(15, 15) × (0.2)¹⁵ × (0.8)⁰ = (0.2)¹⁵. Extremely small: almost no chance every item fails.
P(X < 3): sum the first three terms: P(X = 0) + P(X = 1) + P(X = 2). Each is computed with the formula and then added.
P(X > 2): use the complement: P(X > 2) = 1 – P(X ≤ 2) = 1 – P(X < 3). The complement rule saves work whenever you would otherwise sum many terms.
E(X) = µ = np. For the binding example: 15 × 0.2 = 3. On average, 3 items out of 15 fail.
Var(X) = np(1 – p). For the example: 15 × 0.2 × 0.8 = 2.4.
σ = √[np(1 – p)]. For the example: √2.4 ≈ 1.549.
These shortcuts follow from the general PMF rules covered earlier, but they are much faster than computing E(X) and Var(X) from the full PMF.
P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n\binom{n}{x} = \frac{n!}{x!(n-x)!}E(X) = np\text{Var}(X) = np(1-p)\sigma_X = \sqrt{np(1-p)}Quality control departments use the binomial distribution constantly: if a production line has a known defect rate p, the binomial tells you the probability that a random sample of n items contains x or more defects. Medical trials use it to determine whether a drug's observed success rate is plausibly consistent with a given hypothesis about its true effectiveness.
Students sometimes define "success" as the favourable outcome. It is simply the outcome you are counting, even if that outcome is negative (defective, failed, ill).
Forgetting that n must be fixed before the experiment begins. "Keep testing until you get 5 failures" is not a binomial setup.
Using the binomial when trials are not independent. Drawing cards from a deck without replacement violates independence unless the population is very large relative to the sample.
Mixing up P(X ≤ 2) and P(X < 3). For discrete distributions these are the same thing, but students often treat them differently.
⚠️ You will almost certainly be asked to verify whether a scenario meets all four BINS conditions. Practise writing out each condition and checking it.
⚠️ Complement problems ("more than k") appear frequently. Always consider P(X > k) = 1 – P(X ≤ k) before summing many terms.
⚠️ Know the mean and variance formulas: np and np(1 – p). These are quick marks.
⚠️ Make sure you can compute C(n, x) by hand for small values of n. Calculators help, but the exam may require you to show the factorial expansion.
True or false: in a binomial experiment, "success" must be a positive outcome.
False. Success is just the outcome you are counting.
Fill in the blank: the mean of a binomial distribution is ___.
np.
True or false: C(5, 2) = C(5, 3).
True. C(n, x) = C(n, n – x).
Fill in the blank: P(X > 4) = 1 – P(X ≤ ___).
True or false: if trials are not independent, you can still use the binomial formula.
False. Independence is a required condition.
Q: A machine produces parts with a 10% defect rate. In a sample of 8 parts, what is the probability that exactly 2 are defective?
A: P(X = 2) = C(8, 2) × (0.1)² × (0.9)⁶ = 28 × 0.01 × 0.531441 ≈ 0.1488.
Q: Using the same scenario, what is E(X) and what is σ?
A: E(X) = 8 × 0.1 = 0.8. Var(X) = 8 × 0.1 × 0.9 = 0.72. σ = √0.72 ≈ 0.849.
Q: A batch of 15 items has a 20% failure rate. Is this a binomial experiment? State each condition.
A: Binary (fail/pass), Independent (each item tested separately), n = 15 (fixed), p = 0.2 (constant). All four conditions met; yes, it is binomial.
Q: In the batch of 15, what is P(X < 3)?
A: P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2). Compute each with the binomial formula using n = 15, p = 0.2, then sum.
Q: What is P(X > 2) for the same batch?
A: P(X > 2) = 1 – P(X ≤ 2) = 1 – P(X < 3). Use the complement of the previous answer.
The binomial distribution connects directly to the general PMF and expected-value rules from the first half of Chapter 5. As n gets large and p stays moderate, the binomial can be approximated by the normal distribution (covered in later chapters). When n is large and p is very small, the binomial is well approximated by the Poisson distribution (the next topic in this chapter). The complement rule used here (P(X > k) = 1 – P(X ≤ k)) will reappear in every distribution you study.
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