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 $P(X = 3) = 120\cdot{}0.5^{3}\cdot{}0.5^{7} = 120\cdot{}0.5^{10} =\frac{ 120}{1024} \approx{} 0.117$. 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 $P(X \le{} k)$, which the cumulative distribution function (binomcdf) gives directly. "At least k successes" means $P(X \ge{} k) = 1 - P(X \le{} k - 1)$; the detour via the complement is needed because tables and calculators only know "≤". Careful at the edges: "more than k" means $P(X \ge{} k + 1)$, "fewer than k" means $P(X \le{} k - 1)$. The most popular special case is "at least one success": $P(X \ge{} 1) = 1 - P(X = 0) = 1 - (1-p)^{n}$. Example: at least one six in 10 die rolls: $1 - (\frac{5}{6})^{10} \approx{} 1 - 0.162 = 0.838$. Writing the inequality down before computing prevents the frequent off-by-one errors.

For a binomial random variable, $E(X) = \mu{} = n\cdot{}p$ and $\sigma{} = \sqrt{(n\cdot{}p\cdot{}(1-p))}$. Example: $n = 100$ fair coin tosses give $\mu{} = 50$ and $\sigma{} = 5$. The expectation is the long-run mean count, and σ measures typical deviation. For a normal distribution, μ ± σ contains about 68.27% and μ ± 2σ about 95.45%. For binomial data, however, the bell curve is only an approximation; integer boundaries need checking. Here the exact sum for $40 \le{} X \le{} 60$ is about 96.48%, not 95%. The Laplace condition $\sigma{} > 3$ supports the approximation but does not make it exact. These intervals are common prediction exercises in 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 $\mu{} = np$ and $\sigma{} = \sqrt{(np(1-p))}$; 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.