Key Idea: The binomial distribution counts successes in a fixed number of independent trials — and it is a heavy Paper 2 GDC topic, with binompdf / binomcdf doing nearly all the work.
🎯 When is it binomial? — X ~ B(n, p)
Fixed number of trials n, each trial independent, only two outcomes (success / failure), and the same success probability p every time. Then X ~ B(n, p) counts the successes. Drawing without replacement changes p between trials → not binomial.
🧮 Probabilities on the GDC (Paper 2)
- number of trials
- success probability
- number of successes
| You want… | GDC | How to build it |
|---|---|---|
| P(X = k) (exactly) | binompdf(n, p, k) | Direct. |
| P(X ≤ k) (at most) | binomcdf(n, p, k) | Direct (cumulative). |
| P(X ≥ k) (at least) | binomcdf | 1 − binomcdf(n, p, k − 1) |
| P(X > k) (more than) | binomcdf | 1 − binomcdf(n, p, k) |
| P(a ≤ X ≤ b) | binomcdf | binomcdf(…, b) − binomcdf(…, a − 1) |
| P(at least one) | binompdf | 1 − binompdf(n, p, 0) |
📊 Mean & variance
- expected number of successes
- variance — multiply by both p and (1 − p)
Tip: Given the mean and variance, variance ÷ mean = 1 − p (this cancels n) → gives p; then n = mean ÷ p.
✏️ IB-style worked examples
IB-style question — exactly, at most, at least
A spinner lands on red with probability 0.35. It is spun 14 times. Let X be the number of reds. Find (a) P(X = 5), (b) P(X ≤ 4), (c) P(X ≥ 6).
Step by step:
State the model.
(a) Exactly 5 → binompdf(14, 0.35, 5).
(b) At most 4 → binomcdf(14, 0.35, 4).
(c) At least 6 is the complement of ≤ 5.
(a) 0.218 (b) 0.423 (c) 0.359 (3 s.f.).
IB-style question — find the mean, variance and standard deviation
A quiz has 40 independent questions, each guessed correctly with probability 0.25. Let X be the number correct. Find the mean, variance and standard deviation of X.
Step by step:
Mean = np.
Variance = np(1 − p).
Standard deviation = √variance.
Mean = 10, variance = 7.5, standard deviation ≈ 2.74.
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: binomcdf only gives P(X ≤ k), so an 'at least k' or 'more than k' question needs the complement: P(X ≥ k) = 1 − binomcdf(n, p, k − 1) and P(X > k) = 1 − binomcdf(n, p, k). Forgetting the 1 −, or the k − 1, is the most common lost mark in this topic.
Tap each card to reveal the answer.
X ~ B(20, 0.4): which command for P(X = 8)? binompdf(20, 0.4, 8) — pdf = exactly.
X ~ B(20, 0.4): how do you get P(X ≥ 8)? 1 − binomcdf(20, 0.4, 7) — complement, with k − 1 = 7.
X ~ B(30, 0.2): find the mean and variance Mean = 30 × 0.2 = 6; variance = 30 × 0.2 × 0.8 = 4.8.
A binomial has mean 12 and variance 4.8 — find p variance ÷ mean = 0.4 = 1 − p, so p = 0.6 (then n = 12 ÷ 0.6 = 20).
P(at least one) for n trials, success prob p? 1 − P(X = 0) = 1 − (1 − p)ⁿ — use the complement.
Exam Tips
- Always state the model first: X ~ B(n, p), then read off n, p and k.
- binompdf = exactly k; binomcdf = at most k (≤). They live in DISTR (2nd → VARS).
- 'At least / more than' → use the complement: P(X ≥ k) = 1 − binomcdf(n, p, k − 1).
- Mean = np, variance = np(1 − p), sd = √(np(1 − p)) — all in the formula booklet.
- To find n and p from mean & variance: variance ÷ mean = 1 − p, then n = mean ÷ p.