Quick answer
P(X = k) = C(n, k) p^k (1 − p)^(n − k) for k = 0,…,n in the standard binomial model.
Formula
- C(n,k) counts patterns
- p^k (1−p)^(n−k) weights one pattern
Introduction
Separate counting from weighting. Students often mash them together and lose track of why p appears with exponent k.
Refresh combination counting with worked combination examples if pattern counts feel shaky before you attach p.
Expansion exercises reuse the same coefficients; see binomial theorem and coefficients when your course moves from probability to algebra in the same week.
Where C(n, k) sits in the model
Imagine n independent trials, each success probability p. A specific sequence with k successes and n − k failures has probability p^k (1 − p)^(n − k).
How many distinct success patterns exist? Exactly C(n, k), because you choose which trials succeed without ordering those trials.
Summing over k gives the full distribution. Mean np and variance np(1 − p) come later; the coefficient is the combinatorial backbone.
Binomial probability mass function
- P(X = k) = C(n,k) p^k (1 − p)^(n−k)
Assumptions matter: fixed n, independence, constant p. Drawing cards without replacement from a small deck violates independence; hypergeometric counts replace C(n, k) there.
Step-by-step guide
- Confirm binomial story. Independent trials, two outcomes, fixed n, constant p.
- Count patterns with C(n, k). Use symmetry when k > n/2.
- Attach probability per pattern. Multiply by p^k (1 − p)^(n − k).
- Sum if you need cumulative probabilities. At most k successes sums terms from 0 through k.
Ten trials, three successes
C(10, 3) = 120 patterns.
If p = 0.4, each pattern has weight 0.4^3 × 0.6^7.
P(X = 3) = 120 × 0.4^3 × 0.6^7 ≈ 0.215 (compute carefully on exams).
