Bernoulli Trials: Explained with Examples, Notes and FAQs

A single coin toss is one Bernoulli trial; ten tosses are ten Bernoulli trials. The count of successes X then follows the binomial distribution: P(X = k) = C(n, k)·p^k·q^{n−k}, where q = 1 − p and k runs from 0 to n.

The binomial formula answers how likely any particular count of successes is, while the mean np and variance npq summarise the whole pattern. Quality control charts, election polling and medical trial analysis all start by checking the Bernoulli conditions and then applying this distribution.

Key Concepts

• Four conditions: binary outcomes, constant p, independent trials, fixed n; if any fails, the binomial formula does not apply.
• Success and failure: labels only; success is simply the outcome counted, with P(success) = p and P(failure) = q = 1 − p.
• Binomial PMF: P(X = k) = C(n, k)·p^k·q^{n−k}; the C(n, k) counts the orders of k successes among n trials.
• Mean and variance: E(X) = np and Var(X) = npq, growing linearly with n.
• Shape: symmetric when p = 0.5, right-skewed for small p, left-skewed for large p; approximately normal for large n.
• Limits: sampling without replacement from a small population breaks independence; the hypergeometric distribution replaces the binomial there.

Formulas

Binomial probability: P(X = k) = C(n, k)·p^k·(1 − p)^{n−k} for k = 0, 1, …, n. Mean: E(X) = np. Variance: Var(X) = npq with q = 1 − p. Standard deviation: σ = √(npq). CDF values are sums of PMF terms: P(X ≤ k) = Σ C(n, j)·p^j·q^{n−j} over j = 0 … k. The MGF is (q + p·e^t)^n, from which the mean and variance also follow.

Worked Example

A basketball player makes 40% of free throws, so p = 0.4 and q = 0.6, with independent attempts. In n = 5 attempts, find the probability of exactly 3 makes. Here k = 3 and C(5, 3) = 10. P(X = 3) = 10 × (0.4)³ × (0.6)² = 10 × 0.064 × 0.36 = 10 × 0.02304 = 0.2304. The expected makes are np = 5 × 0.4 = 2, with variance npq = 5 × 0.4 × 0.6 = 1.2 and standard deviation √1.2 ≈ 1.095. So 3 makes is a little above average and quite plausible at about 23%, while all 5 makes would be (0.4)^5 ≈ 0.0102, about 1%.

Common Misconceptions

• p can change between trials: Not in Bernoulli trials. If the success probability drifts, for example a tiring player, the trials are not Bernoulli and the binomial formula is invalid.
• Exactly k means at least k: P(X = 3) = 0.2304 above is not P(X ≥ 3); at least 3 needs the sum P(3) + P(4) + P(5). Read the wording precisely.
• The most likely outcome equals the mean: Close but not identical. The mode is floor((n+1)p); here floor(6 × 0.4) = 2, matching the mean 2 this time, but rounding can separate them.

Real-World Applications

Pollsters treat each respondent as a trial to estimate vote shares with margins of error from npq. Reliability engineers count component failures and geneticists count offspring showing a trait, all through the binomial distribution.

Bernoulli trials are n independent repetitions of a success-or-failure experiment with constant probability p, and the success count follows the binomial PMF C(n, k)·p^k·q^{n−k} with mean np and variance npq. The four conditions, binary outcomes, constant p, independence and fixed n, must all hold. From free throws to factory defects to clinical trials, this is the standard model of counted successes.