Five Settings, No Formula

First, a quick retrieval check — different concept, new setting.

A startup’s traffic comes from four channels whose shares must sum to 1 (Lesson 12’s rule: every answer sheet of plausibilities has to add up to one, whether it’s a die’s faces or something else entirely). Organic search: 35%. Paid ads: 25%. Email: 15%. Referral is the remaining share. What’s referral’s share?

(Confirmed in the solution.)

Lessons 15 and 20 gave you two conjugate pairs — beta+binomial, normal+normal — where the posterior update collapses to arithmetic on a couple of parameters. That’s a huge convenience, but it depends on your actual prior belief genuinely belonging to the matching family. Sometimes it just doesn’t.

Scenario. A device can only be manufactured at one of exactly five known factory settings, producing a defect rate p of 0.1, 0.3, 0.5, 0.7, or 0.9 — nothing in between is physically possible. Based on which factories are currently running this product line, your prior belief across those five settings is:

This is not a beta distribution — it has five-point discrete support, not a smooth curve over [0,1] — so none of the beta-binomial machinery applies. But Bayes’ rule doesn’t care whether your prior is smooth: it works on any prior, by brute force if necessary.

Grid approximation, the mechanism:

1. Take each candidate value of p and its prior weight.
2. Multiply by the likelihood of the observed data at that value of p (the same binomial likelihood formula from Lesson 13: C(n,k) p^k (1−p)^(n−k) — or since C(n,k) is the same constant across every row here, you can drop it and use p^k(1−p)^(n−k) directly, same reasoning as Lesson 16).
3. Normalize (divide every row by the total) so the weights sum back to 1 — that’s your posterior, one number per grid point.

Data: you observe 3 successes in 4 trials.

Part 1 — Compute the likelihood at each grid point. For each of the 5 candidate p values, compute p³(1−p) (dropping the shared C(4,3)=4 constant, or keep it — it cancels in the normalization either way; try it without first).

Part 2 — Unnormalized posterior. Multiply each prior weight by its likelihood. Sum all five products.

Part 3 — Normalize. Divide each unnormalized value by the sum from Part 2. Confirm the five normalized values sum to 1.

Part 4 — Posterior mean (the numeric answer above). Compute Σ p × (normalized posterior weight) across all five grid points. How does the data (3 out of 4 successes, a 75% raw rate) shift the distribution compared to the prior, which was heavily weighted toward p = 0.5?