Bayes' Rule Practice
You test positive for a rare disease. How worried should you be?
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Question 1 · Easy · medical test
A disease affects 1% of people. A test catches 95% of sick people and wrongly flags 5% of healthy people. You test positive. What is the probability you have the disease?
Say your approach out loud first.
Answer: about 16%
- Imagine 10,000 people: 100 are sick, 9,900 healthy.
- Sick and positive: 95% of 100 = 95.
- Healthy but positive: 5% of 9,900 = 495.
- P(sick | positive) = 95 / (95 + 495) = 95/590 ≈ 0.161.
Follow-up: What if you test positive twice (independent tests)? Use 0.161 as the new prior: 0.161×0.95 / (0.161×0.95 + 0.839×0.05) ≈ 0.785.
Question 2 · Easy · coins
A bag has two coins: one fair, one with heads on both sides. You pick one at random, flip it, and get heads. What is the probability it's the double-headed coin?
Say your approach out loud first.
Answer: 2/3
- P(heads | double) = 1, P(heads | fair) = 1/2, each coin picked with probability 1/2.
- P(double | heads) = (½ × 1) / (½ × 1 + ½ × ½) = (½) / (¾) = 2/3.
Follow-up: You flip the same coin again and get heads again. Now? (½·1)/(½·1 + ½·¼) = 4/5.
Question 3 · Medium · witness
In a city 85% of taxis are green and 15% are blue. A witness says the taxi in an accident was blue. Tested in the same conditions, the witness identifies colours correctly 80% of the time. What is the probability the taxi was blue?
Say your approach out loud first.
Answer: about 41%
- Of 100 taxis: 15 blue, 85 green.
- Witness says "blue": correct for 80% of the 15 blue = 12; wrong for 20% of the 85 green = 17.
- P(blue | says blue) = 12 / (12 + 17) = 12/29 ≈ 0.414.
Follow-up: Why is it below 50% even though the witness is 80% reliable? The base rate (few blue taxis) dominates: the classic base-rate fallacy (Kahneman & Tversky).
Question 4 · Medium · two children
A family has two children. You learn that at least one of them is a boy. What is the probability both are boys? (Assume boys and girls are equally likely and independent.)
Say your approach out loud first.
Answer: 1/3
- Equally likely families: BB, BG, GB, GG.
- "At least one boy" rules out GG, leaving BB, BG, GB.
- Only BB has two boys: 1/3.
Follow-up: If instead you learn the older child is a boy, the answer is 1/2 (only BB and BG remain). The wording changes the conditioning event.
Question 5 · Medium · spam filter
40% of emails are spam. The word "free" appears in 30% of spam emails and in 2% of normal emails. An email contains "free". What is the probability it's spam?
Say your approach out loud first.
Answer: about 91%
- Of 1,000 emails: 400 spam, 600 normal.
- Contain "free": 30% of 400 = 120 spam; 2% of 600 = 12 normal.
- P(spam | free) = 120 / 132 ≈ 0.909.
Follow-up: How does Naive Bayes combine several words? Multiply each word's likelihood ratio, assuming the words are independent given the class.
Question 6 · Hard · boxes
Three boxes: one has two gold coins, one has two silver, one has one of each. You pick a box at random and draw a coin: it's gold. What is the probability the other coin in that box is also gold?
Say your approach out loud first.
Answer: 2/3
- Count gold coins you could have drawn: 2 in the gold-gold box, 1 in the mixed box: 3 equally likely gold coins.
- In 2 of those 3 cases you're in the gold-gold box, so the other coin is gold.
- P = 2/3, not 1/2 (Bertrand's box paradox).
Follow-up: Why is "1/2" tempting? It treats the two remaining boxes as equally likely, ignoring that the gold-gold box is twice as likely to produce a gold draw.
More practice and sources
6 interview questions, easy to hard · answers hidden until you click · checked 28 Oct 2026
These are classic interview questions. For more in the same style: Frederick Mosteller, Fifty Challenging Problems in Probability with Solutions (Dover, 1987), and Xinfeng Zhou, A Practical Guide to Quantitative Finance Interviews (2008). Related lessons: Naive Bayes, Classification Metrics.