Binomial Distribution

The experiment must be a Bernoulli chain, meaning four conditions hold: there is a fixed number n of trials; each trial has exactly two outcomes (success/failure); the success probability p is the same in every trial; the trials are independent. The random variable X then counts the successes. Classic examples: coin tosses, rolling a die for a specific number, guessing on multiple choice. Drawing without replacement, such as lots from an urn, is critical: there p changes from draw to draw, and the hypergeometric distribution is exactly responsible. But if the population is very large compared to the sample, p barely changes and the binomial distribution is a good approximation; this is exactly how many exam tasks argue.

Insert n, k and p into P(X = k) = (n choose k)·pᵏ·(1−p)ⁿ⁻ᵏ. The binomial coefficient (n choose k) counts in how many ways the k successes can be distributed over the n positions; pᵏ·(1−p)ⁿ⁻ᵏ is the probability of one such path. Example: 10 coin tosses, exactly 3 heads: (10 choose 3) = 120, so P(X = 3) = 120·0.5³·0.5⁷ = 120·0.5¹⁰ = 120/1024 ≈ 0.117. On a calculator the function binompdf(n, p, k) or BinomialPD does this. Make sure to assign p and k correctly: p belongs to the exponent of the successes, 1 − p to the failures, and both exponents must add up to n.

First translate the words into inequalities. "At most k successes" means P(X ≤ k), which the cumulative distribution function (binomcdf) gives directly. "At least k successes" means P(X ≥ k) = 1 − P(X ≤ k − 1); the detour via the complement is needed because tables and calculators only know "≤". Careful at the edges: "more than k" means P(X ≥ k + 1), "fewer than k" means P(X ≤ k − 1). The most popular special case is "at least one success": P(X ≥ 1) = 1 − P(X = 0) = 1 − (1−p)ⁿ. Example: at least one six in 10 die rolls: 1 − (5/6)¹⁰ ≈ 1 − 0.162 = 0.838. Writing the inequality down before computing prevents the frequent off-by-one errors.

For a binomially distributed random variable the short formulas E(X) = μ = n·p and σ = √(n·p·(1−p)) hold. Example: n = 100 tosses of a fair coin give μ = 50 and σ = √(100·0.5·0.5) = 5. The expected value is the long-run average number of successes, the standard deviation measures the typical deviation from it. Together they carry the sigma rules: about 68% of outcomes lie in [μ − σ; μ + σ], about 95% in [μ − 2σ; μ + 2σ], provided the Laplace condition σ > 3 is met. In the example this means: with about 95% probability between 40 and 60 heads occur. Such prediction intervals are a standard component of final exams.

The binomial distribution is discrete: it assigns individual probabilities to the integer success counts k = 0, 1, ..., n and fits counting experiments. The normal distribution is continuous: it describes quantities like measurements via a density function, and probabilities are areas under the bell curve. The bridge: for large n the histogram of the binomial distribution itself looks like a bell curve, and it may be approximated by a normal distribution with the same μ = np and σ = √(np(1−p)); as a rule of thumb one requires σ > 3 (Laplace condition). This approximation underlies the sigma rules of stochastics. So in the exam you count discretely with the binomial distribution and use normal-distribution logic for interval statements.