A discrete probability distribution used to model the number of successes in a fixed number of independent trials, each with constant probability of success.
## Core concept
The binomial distribution describes outcomes when: - Fixed number of trials (n) are conducted - Each trial has exactly two outcomes: success or failure - Probability of success (p) remains constant across all trials - Trials are independent of each other
Notation: X ~ B(n, p), where n = number of trials, p = probability of success per trial.
Real exam contexts: Quality control sampling, pass/fail exam results, defective items in production batches, coin tosses, yes/no survey responses.
## Formula / rule
Probability Mass Function (PMF):
P(X = r) = C(n, r) × p^r × (1 − p)^(n−r)
where: - C(n, r) = nCr = n! / [r!(n−r)!] (combinations) - r = number of successes (0 ≤ r ≤ n) - p = probability of success - (1 − p) = q = probability of failure
Key parameters: - Mean (Expected value): E(X) = np - Variance: Var(X) = npq = np(1 − p) - Standard deviation: σ = √[np(1 − p)]
## Common exam applications
Type 1: Single probability calculation - Calculate P(X = r) for specific r - Calculate P(X ≤ r) or P(X ≥ r) using cumulative approach - Use combination tables or calculators for C(n, r)
Type 2: Mean and variance - Given n and p, find E(X) and Var(X) - Compare two binomial distributions by their parameters
Type 3: Word problem recognition - Identify whether a scenario fits binomial model (check: fixed n? two outcomes? constant p? independent trials?) - Distinguish from Poisson (no fixed n) and Normal (continuous data)
## Worked example
Question: A quality inspector finds that 10% of items in a batch are defective. If 5 items are randomly selected, find: (a) Probability that exactly 2 items are defective (b) Expected number of defective items
Solution: - n = 5, p = 0.10, q = 0.90
(a) P(X = 2) = C(5, 2) × (0.10)² × (0.90)³ - C(5, 2) = 5!/(2!3!) = 10 - P(X = 2) = 10 × 0.01 × 0.729 = 0.0729 or 7.29%
(b) E(X) = np = 5 × 0.10 = 0.5 items (expected defectives)
## Common mistakes
- Confusing p and q: Always check which is success probability
- Forgetting combination term C(n, r): PMF requires all three components; probability alone is incomplete
- Using binomial when n is not fixed: If sampling without replacement or n is variable, consider hypergeometric or Poisson instead
- Misinterpreting "at most" vs "at least": P(X ≤ r) ≠ P(X ≥ r); cumulative sums differ
- Rounding errors in tables: Exam answers often require 4 decimal places; preserve precision until final step
- Assuming independence incorrectly: Sampling without replacement violates binomial assumption; use hypergeometric distribution instead
Exam tip: Always state the assumptions clearly (fixed n, two outcomes, constant p, independence) to justify using binomial distribution—examiners check this reasoning.