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Probability & Statistics · Probability Foundations

Bayes' Rule — Practice Problems and Worked Solutions

Turn prior beliefs and evidence into posterior probability.

The formulas you need here

Bayes’ rule

P (A ∣ B) = \frac{P ( B ∣ A ) P ( A )}{P ( B )}

What it says: Turn a conditional probability around using the prior.
When to reach for it: Medical tests, spam filters, and any "given a positive result, what is the chance" question.
Watch out: The denominator usually needs the law of total probability: P(B) = P(B|A)P(A) + P(B|A′)P(A′). A rare condition keeps P(A|B) low even after a positive test.

P (B) = P (B ∣ A) P (A) + P (B ∣ A^{c}) P (A^{c})

Compute the denominator first — it is where the marks are lost.

16 worked problems

Every step is generated and checked by a computer algebra system.

1.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{10}, P (+ ∣ H) = \frac{4}{5}, P (+ ∣ H^{c}) = \frac{1}{10}$
Answer: $\frac{8}{17}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{8}{17}$
2.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{5}, P (+ ∣ H) = \frac{3}{4}, P (+ ∣ H^{c}) = \frac{1}{5}$
Answer: $\frac{15}{31}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{15}{31}$
3.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{4}, P (+ ∣ H) = \frac{2}{3}, P (+ ∣ H^{c}) = \frac{1}{6}$
Answer: $\frac{4}{7}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{4}{7}$
4.
Find the posterior probability P(H|+).
$P (H) = \frac{2}{5}, P (+ ∣ H) = \frac{3}{5}, P (+ ∣ H^{c}) = \frac{1}{5}$
Answer: $\frac{2}{3}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{2}{3}$
5.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{20}, P (+ ∣ H) = \frac{9}{10}, P (+ ∣ H^{c}) = \frac{1}{20}$
Answer: $\frac{18}{37}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{18}{37}$
6.
Find the posterior probability P(H|+).
$P (H) = \frac{3}{10}, P (+ ∣ H) = \frac{4}{5}, P (+ ∣ H^{c}) = \frac{1}{10}$
Answer: $\frac{24}{31}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{24}{31}$
7.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{8}, P (+ ∣ H) = \frac{4}{5}, P (+ ∣ H^{c}) = \frac{1}{8}$
Answer: $\frac{32}{67}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{32}{67}$
8.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{3}, P (+ ∣ H) = \frac{3}{4}, P (+ ∣ H^{c}) = \frac{1}{4}$
Answer: $\frac{3}{5}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{3}{5}$
9.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{10}, P (+ ∣ H) = \frac{9}{10}, P (+ ∣ H^{c}) = \frac{1}{5}$
Answer: $\frac{1}{3}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{1}{3}$
10.
Find the posterior probability P(H|+).
$P (H) = \frac{2}{5}, P (+ ∣ H) = \frac{1}{2}, P (+ ∣ H^{c}) = \frac{1}{10}$
Answer: $\frac{10}{13}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{10}{13}$
11.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{6}, P (+ ∣ H) = \frac{2}{3}, P (+ ∣ H^{c}) = \frac{1}{6}$
Answer: $\frac{4}{9}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{4}{9}$
12.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{4}, P (+ ∣ H) = \frac{4}{5}, P (+ ∣ H^{c}) = \frac{1}{5}$
Answer: $\frac{4}{7}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{4}{7}$
13.
Find the posterior probability P(H|+).
$P (H) = \frac{3}{5}, P (+ ∣ H) = \frac{2}{3}, P (+ ∣ H^{c}) = \frac{1}{3}$
Answer: $\frac{3}{4}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{3}{4}$
14.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{5}, P (+ ∣ H) = \frac{1}{2}, P (+ ∣ H^{c}) = \frac{1}{4}$
Answer: $\frac{1}{3}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{1}{3}$
15.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{2}, P (+ ∣ H) = \frac{3}{5}, P (+ ∣ H^{c}) = \frac{1}{5}$
Answer: $\frac{3}{4}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{3}{4}$
16.
Find the posterior probability P(H|+).
$P (H) = \frac{1}{20}, P (+ ∣ H) = \frac{4}{5}, P (+ ∣ H^{c}) = \frac{1}{10}$
Answer: $\frac{8}{27}$
Step-by-step solution
1. $P (+) = P (+ ∣ H) P (H) + P (+ ∣ H^{c}) P (H^{c})$
2. $P (H ∣ +) = \frac{8}{27}$

The formulas you need here

Bayes’ rule

16 worked problems

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