Conditional & Bayes: interview questions and key results

Updating beliefs from evidence. The results to know for quant trading interviews, then a worked example of each of the 5 question types. Free, no sign-up.

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Key results

Worked examples (5)

Try each one before opening the solution. In the app, every type generates fresh numbers each time.

Base rates (medical test)

A condition affects 5% of people. A test detects it 95% of the time when present and gives a false positive 2% of the time when absent. Someone tests positive. What is the probability they have the condition?

Answer and worked solution

Answer: 5/7 ≈ 0.71429

P(D|+) = P(+|D)P(D) / [P(+|D)P(D) + P(+|¬D)P(¬D)] = (0.95·0.05) / (0.95·0.05 + 0.02·0.95) ≈ 0.71429. With a low base rate, most positives are false.

Odds-form Bayes

A bag has 10 coins: 9 fair and 1 double-headed. You pick one at random and flip it 6 times, getting 6 heads. What is the probability you picked the double-headed coin?

Answer and worked solution

Answer: 0.87671

Prior odds (double : fair) = 1 : 9. Likelihood ratio = 1 / (1/2)6 = 64. Posterior odds = 64 : 9, so P = 64/73 ≈ 0.87671. Thinking in odds makes Bayes fast.

Conditioning on "at least one"

A family has 4 children. Given at least one is a boy, what is the probability all 4 are boys? (Each child is independently a boy or girl with probability 1/2.)

Answer and worked solution

Answer: 1/15 ≈ 0.066667

Of the 16 equally likely sequences, 15 contain at least one boy and one is all boys. P = 1/15 ≈ 0.066667.

Conditioning on dice sums

Two dice are rolled and you are told the sum is at least 7. What is the probability at least one die shows a six?

Answer and worked solution

Answer: 11/21 ≈ 0.52381

Restrict to the 21 outcomes with sum ≥ 7. Of these, 11 contain a six. P = 11/21 ≈ 0.52381.

Two-urn Bayes

Urn A has 3 red and 4 blue balls; urn B has 6 red and 5 blue. You pick an urn with a fair coin and draw one ball: it's red. What is the probability it came from urn A?

Answer and worked solution

Answer: 0.44

P(A|red) = P(red|A)·½ / [P(red|A)·½ + P(red|B)·½] = 0.42857 / (0.42857 + 0.54545) ≈ 0.44.

Other topics

Dice & coins · Cards & combinatorics · Expected value & games · Random walks & martingales · Markov chains & order statistics · Options · Classic brainteasers · Number sequences · Distributions · Variance & covariance · Estimation & testing · Regression & correlation · Returns, risk & sizing · Time series

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