Key Idea: The binomial distribution models a situation where you repeat the same experiment n times independently, each time with the same probability p of success. It counts how many successes you get in total. The key is checking the four conditions: fixed n, constant p, independent trials, binary outcome (success/fail).
✅ Binomial conditions and notation
Example: X ~ B(8, 0.4) P(X = 3) = binompdf(8, 0.4, 3) = 0.279 (3 s.f.) P(X ≤ 3) = binomcdf(8, 0.4, 3) = 0.594 P(X ≥ 4) = 1 − binomcdf(8, 0.4, 3) = 1 − 0.594 = 0.406 P(2 ≤ X ≤ 5) = binomcdf(8, 0.4, 5) − binomcdf(8, 0.4, 1) = 0.950 − 0.106 = 0.844 E(X) = 8 × 0.4 = 3.2
The most common error: using binomcdf(n, p, k) for P(X ≥ k) instead of 1 − binomcdf(n, p, k−1). The '−1' is critical when working with 'at least k'. Before using binomial: check all four conditions in your working. If trials are not independent or p is not constant, the binomial model does not apply.
Paper 2 (GDC allowed): Always write the distribution statement X ~ B(n, p) first, then write the probability statement (e.g., P(X ≥ 4)), then calculate. This structure earns communication marks. Paper 1: You may need to use the formula P(X = k) = C(n,k) × pᵏ × (1−p)ⁿ⁻ᵏ for small n. Know how to calculate C(n,k) = n!/(k!(n−k)!).
IB-style question [6 marks]
An online shop knows that, independently, 20% of all orders are returned. On a particular day it receives 15 orders. Let X be the number of these orders that are returned. (a) Write down the expected number of returned orders. (b) Find the probability that exactly 2 orders are returned. (c) Find the probability that at least 4 orders are returned.
Step by step:
Each order is independent with a constant return probability, so X is binomial with n = 15 and p = 0.2.
(a) The expected number is the binomial mean np.
(b) 'Exactly 2' uses the single-value probability binompdf.
(c) 'At least 4' is the complement of 'at most 3', so subtract the cumulative probability up to 3.
Use binomcdf(15, 0.2, 3), then subtract from 1.
(a) 3 orders. (b) 0.231. (c) 0.352.