Chapter 34 — Bayes' Theorem Challenge Solutions

Chapter 34 · Solutions

Bayes' Theorem, Worked Answers ✅

Full solutions to the five challenges. Try them yourself first, then compare.

Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026

⚙️ Setup¶

In [1]:

import numpy as np
rng = np.random.default_rng(341)
print("ready")

ready

CHALLENGE 1 · REVERSE A CONDITIONAL

Bayes from a tree

70% of email is legitimate, 30% is spam. The word 'offer' appears in 60% of spam and 5% of legit mail. An email contains 'offer'. Find P(spam | 'offer').

In [2]:

p_spam, p_legit = 0.30, 0.70
p_off_spam, p_off_legit = 0.60, 0.05
p_off = p_spam*p_off_spam + p_legit*p_off_legit     # total probability
post  = p_spam*p_off_spam / p_off                    # Bayes
print(f"P(\"offer\")           = {p_off:.3f}")
print(f"P(spam | \"offer\")    = {post:.3f}")

P("offer")           = 0.215
P(spam | "offer")    = 0.837

Answer: P("offer") = 0.30(0.60) + 0.70(0.05) = 0.18 + 0.035 = 0.215. Then P(spam | "offer") = 0.18 / 0.215 ≈ 0.837. One suggestive word lifts the spam probability from a 0.30 prior to about 84%.

CHALLENGE 2 · LAW OF TOTAL PROBABILITY

Average over three suppliers

Three suppliers send parts: S1 = 50% of stock at 1% defective, S2 = 30% at 2%, S3 = 20% at 3%. What fraction of all parts are defective?

In [3]:

w   = np.array([0.50, 0.30, 0.20])     # share from each supplier
d   = np.array([0.01, 0.02, 0.03])     # defect rate of each
p_def = (w*d).sum()
for i,(wi,di) in enumerate(zip(w,d),1):
    print(f"S{i}: {wi:.2f} * {di:.2f} = {wi*di:.3f}")
print(f"P(defective) = {p_def:.3f}")

S1: 0.50 * 0.01 = 0.005
S2: 0.30 * 0.02 = 0.006
S3: 0.20 * 0.03 = 0.006
P(defective) = 0.017

Answer: the Law of Total Probability averages the defect rates, weighted by each supplier's share: P(defective) = 0.50(0.01) + 0.30(0.02) + 0.20(0.03) = 0.005 + 0.006 + 0.006 = 0.017, that is 1.7%. The suppliers partition every part, so the slices simply add.

CHALLENGE 3 · PRIOR → POSTERIOR

A single test

A condition has 2% prevalence. A test has 90% sensitivity and 85% specificity. Given one positive result, find P(condition | positive).

In [4]:

prior, sens, spec = 0.02, 0.90, 0.85
p_pos = prior*sens + (1-prior)*(1-spec)
post  = prior*sens / p_pos
print(f"P(positive)             = {p_pos:.3f}")
print(f"P(condition | positive) = {post:.3f}  (about {post:.0%})")

P(positive)             = 0.165
P(condition | positive) = 0.109  (about 11%)

Answer: P(positive) = 0.02(0.90) + 0.98(0.15) = 0.018 + 0.147 = 0.165. Posterior = 0.018 / 0.165 ≈ 0.109, about 11%. Even a positive test leaves the condition unlikely, because a 15% false-positive rate on the large healthy group produces many more false alarms than true ones.

CHALLENGE 4 · SEQUENTIAL UPDATING

A second positive test

The patient from Challenge 3 takes a second, independent test that is also positive. Update again, using the previous posterior as the new prior.

In [5]:

prior = 0.109                 # the posterior from Challenge 3
sens, spec = 0.90, 0.85
p_pos = prior*sens + (1-prior)*(1-spec)
post  = prior*sens / p_pos
print(f"new prior (was posterior) = {prior:.3f}")
print(f"P(condition | 2 positives) = {post:.3f}  (about {post:.0%})")

new prior (was posterior) = 0.109
P(condition | 2 positives) = 0.423  (about 42%)

Answer: with the new prior 0.109: P(positive) = 0.109(0.90) + 0.891(0.15) = 0.0981 + 0.1337 = 0.2318, so the posterior = 0.0981 / 0.2318 ≈ 0.423, about 42%. Two positives are far more convincing than one: belief climbed from 2% to 11% to 42% as independent evidence accumulated.

CHALLENGE 5 · NAIVE BAYES

P(class | feature)

A spam filter has prior P(spam) = 0.5. The word 'win' appears in 70% of spam and 5% of ham. Given 'win', find P(spam | 'win').

In [6]:

p_spam = 0.50
p_win_spam, p_win_ham = 0.70, 0.05
p_win = p_spam*p_win_spam + (1-p_spam)*p_win_ham
post  = p_spam*p_win_spam / p_win
print(f"P(\"win\")          = {p_win:.3f}")
print(f"P(spam | \"win\")   = {post:.3f}  (about {post:.0%})")

P("win")          = 0.375
P(spam | "win")   = 0.933  (about 93%)

Answer: P("win") = 0.5(0.70) + 0.5(0.05) = 0.35 + 0.025 = 0.375. Posterior = 0.35 / 0.375 ≈ 0.933, about 93%. The strong likelihood ratio (0.70 vs 0.05) makes "win" a powerful spam signal; a full classifier multiplies many such word likelihoods together.

🎉 Nicely done

You reversed conditionals with Bayes, averaged over a partition with the Law of Total Probability, updated a prior into a posterior, chained two updates as evidence accumulated, and scored a class with naive Bayes. The pattern never changed: posterior ∝ likelihood × prior.