← Apply probability distributions

6 Combining probabilities across stages

The Excellence part of a question is almost always this: taking a probability you have already found and doing something further with it — repeating it across independent trials, multiplying two different distributions together, or finding a conditional probability from one model.

Three ways probabilities combine

  • Repeated independent trials of the same event: if event A happens independently at each of n identical stores (or days, or batches), P(A at all n) = P(A)ⁿ.
  • Two DIFFERENT distributions in one scenario: if the events are independent, multiply the two probabilities directly — one from a binomial, one from a normal, one from a triangular, it does not matter which, as long as they are independent.
  • Conditional probability from a single continuous model: P(X > b | X > a) = P(X > b) / P(X > a) for a < b — exactly the same conditional-probability rule as tree diagrams, just with areas instead of branches.
Worked example Multiplying across independent events A company estimates P(at least 3 of the next 10 customers make a purchase) = 0.7384 at one store. Assuming independence between stores, estimate the probability this happens at all 4 of the company's stores. The four stores are independent, so multiply the SAME probability across all four (not four different probabilities). P(all 4 stores) = 0.7384⁴. 0.7384⁴ ≈ 0.2973. ✓ P(all 4 stores) ≈ 0.297. Note the assumption this relies on: that the 0.35 purchase rate genuinely holds identically at every store, which is worth challenging in context (richer or poorer areas may differ). Worked example Conditional probability from a triangular model Customer spend is triangular with a = 0, b = 250, c = 130. Given that a customer spends more than
30, find the probability they spend more than
50. This is conditional: P(X > 150 | X > 130) = P(X > 150) / P(X > 130). Both 130 and 150 sit on the falling side (c = 130 to b = 250), so use P(X > x) = (b − x)²/[(b − a)(b − c)]. P(X > 130) = (250 − 130)²/[(250)(120)] = 14400/30000 = 0.48. P(X > 150) = (250 − 150)²/[(250)(120)] = 10000/30000 = 0.3333. P(X > 150 | X > 130) = 0.3333 / 0.48 ≈ 0.6944. ✓ P(X > 150 | X > 130) ≈ 0.694. Sanity check: this conditional probability must be LARGER than the unconditional P(X > 150) = 0.333, because you already know the customer spent more than
30 — and 0.694 > 0.333. ✓ ✕

A repeated-trial probability is a power, not a sum or a product of different numbers

When the SAME independent event must happen at every one of n identical stages, the answer is P(event)n — a single probability raised to a power. A common error is instead adding the probability n times, or multiplying together n DIFFERENT-looking numbers that are actually all the same quantity restated. Before combining anything, ask: are these genuinely independent, and is it really the same event repeating, or two different events that both need to occur?

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