A quant probability interview guide with 25 original questions, answers and explanations covering expectation, Bayes, counting and stochastic processes.
An attractive expected payoff can hide a bad conditioning argument or an unjustified stopping rule. These questions train the setup before the arithmetic: define the random experiment, identify the information available and check the result against a simple limiting case. Start with indicators and Bayes, then move to overlapping patterns and processes. The three focused collections below give you a longer workout in the areas most heavily represented in our probability bank.
Specify whether draws use replacement, what is observed and whether variables are independent. A change in the observation protocol can change the answer even when the story sounds identical.
Try an indicator for a count, total expectation for a mixture and a state recurrence for a waiting problem. A carefully chosen state often removes pages of enumeration.
Probabilities must lie between zero and one, and waiting times must have the right units. Test symmetry and boundary cases before committing to the result.
Try each question before reading its answer. These original teaching examples remain free to read on this page.
Separate the expected count, gross payout and net profit. Linearity of expectation remains useful when the underlying outcomes are dependent.
Each of 12 independent trades wins with probability 0.55. What is the expected number of wins?
Answer: 6.6 wins.
Explanation: Write the count as a sum of 12 indicators. Each has expectation 0.55; add them. Independence is unnecessary for this expectation.
Twenty signals each fire with probability 0.3, but may be dependent. What is the expected number firing?
Answer: 6 signals.
Explanation: Linearity of expectation needs finite expectations, not independence. Dependence changes the distribution and variance of the total.
A hypothetical trade earns 8 units with probability 0.4 and loses 3 otherwise. A fixed cost of 0.5 applies. Find expected net payoff.
Answer: 0.9 units.
Explanation: Compute 0.4 × 8 − 0.6 × 3 − 0.5. Subtract the cost on every trade, not just winning trades.
A fair six-sided die pays the square of its face value. What is the fair entry price?
Answer: 91/6 units.
Explanation: Average the six payoffs: (1 + 4 + 9 + 16 + 25 + 36)/6. Squaring the average face value gives a different quantity.
A ticket costs 4 units and pays 100 with probability 0.03, zero otherwise. What is its expected profit?
Answer: −1 unit.
Explanation: The expected payout is 3. Subtract the certain entry cost; a large possible payout does not imply positive expected profit.
Use geometric restarts for repeated independent attempts and a distribution or tail integral for minima and maxima.
Independent order attempts each fill with probability 0.2. What is the expected number of attempts through the first fill?
Answer: 5 attempts.
Explanation: The geometric waiting time counts the successful attempt and has mean 1/p. Counting only failures would give 4.
Independent trials succeed with probability 0.25. What is the expected number of trials through the second success?
Answer: 8 trials.
Explanation: Split the wait into two geometric waiting times. Each has mean 4, and expectations add.
Roll a fair die; if it shows 1 or 2, restart under the same rule. Otherwise accept the face value. Find expected accepted score.
Answer: 4.5.
Explanation: Let E be the value. Then E = (2/6)E + (3 + 4 + 5 + 6)/6. Solve; the accepted scores are uniform on 3 through 6.
Two independent quotes are uniform on [0,1]. What is the expected larger quote?
Answer: 2/3.
Explanation: The maximum has CDF x² on [0,1]. Integrating its survival probability gives the mean: integral of 1 − x².
Two independent delays are uniform on [0,1]. What is the expected smaller delay?
Answer: 1/3.
Explanation: The minimum exceeds x only if both delays exceed x, with probability (1 − x)². Integrate from 0 to 1.
Track base rates and the likelihood of the actual evidence. Multiple observations require a defensible dependence model.
A fictional fault occurs in 10% of cases. An alert fires for 80% of faults and 20% of non-faults. Given an alert, what is fault probability?
Answer: 4/13, approximately 30.8%.
Explanation: The numerator is 0.1 × 0.8. Divide by 0.1 × 0.8 + 0.9 × 0.2; false alerts outnumber true alerts.
Use the same 10% fault prior, 80% sensitivity and 20% false-alert rate. Given no alert, what is fault probability?
Answer: 1/37, approximately 2.7%.
Explanation: No-alert likelihood is 0.2 under a fault and 0.8 otherwise. Bayes requires the likelihood of the evidence actually observed.
A fault has prior probability 0.1. Two alerts are conditionally independent given fault status, each with sensitivity 0.8 and false-positive rate 0.2. Given both alerts, find fault probability.
Answer: 0.64/(0.64 + 0.36) = 0.64.
Explanation: Square the two likelihoods within each status, then weight by the prior. Conditional independence must be stated before multiplying.
Two screens show exactly the same alert from one detector. Does observing both give more evidence than observing either?
Answer: No.
Explanation: The second screen is a deterministic copy of the first. Treating the displays as independent would count the same evidence twice.
Prior odds for regime A versus B are 1:4. Observed evidence has likelihood ratio 6 in favour of A. Find posterior probability of A.
Answer: 3/5.
Explanation: Posterior odds are 6:4, or 3:2. Convert odds to probability by dividing the first number by the total.
List the outcomes that remain possible after the observation. The denominator must describe the stated information.
Roll two independent fair dice. Given that at least one shows 6, what is the probability both show 6?
Answer: 1/11.
Explanation: Eleven ordered pairs contain at least one 6. Only one is (6,6); the conditioning event is not a named die being 6.
Roll two independent fair dice. Given that the first die shows 6, what is the probability both show 6?
Answer: 1/6.
Explanation: The named observation fixes only the first die. Compare this sampling protocol with conditioning on at least one six.
Two independent fair dice sum to 8. What is the probability the first die is 3?
Answer: 1/5.
Explanation: The feasible ordered pairs are (2,6), (3,5), (4,4), (5,3), (6,2). They remain equally likely after conditioning.
A bag contains 4 red and 6 blue tokens. Given the first draw is red and is not replaced, what is the chance the next is red?
Answer: 1/3.
Explanation: Three red tokens remain among nine. The first observation changes the composition of the next draw.
Bag A has 3 red and 1 blue token; bag B has 1 red and 3 blue. Choose a bag uniformly and draw a red token. Find probability the bag is A.
Answer: 3/4.
Explanation: The red likelihoods are 3/4 and 1/4, and the priors are equal. Normalise their weighted likelihoods.
These examples add overlapping patterns, uneven arrival probabilities or endpoint conditioning. Write the state equations before simplifying.
Independent fair coin flips continue until HTH first appears. What is the expected number of flips?
Answer: 10.
Explanation: Track matched prefixes: empty, H, HT. The equations E₀ = 2 + E_H, E_H = 2 + E_HT and E_HT = 1 + E₀/2 give E₀ = 10.
With independent fair coin flips, what is the probability HHT appears before HTH?
Answer: 2/3.
Explanation: After the first H, a second H leads to HH, which eventually completes HHT. From HT, a head loses and a tail restarts.
Independent labels A, B, C occur with probabilities 1/2, 1/3, 1/6. Find expected draws to collect all three.
Answer: 73/10.
Explanation: Inclusion–exclusion of missing-label tail probabilities gives 2 + 3 + 6 − 6/5 − 3/2 − 2 + 1.
A walk steps up with probability 2/3, down with 1/3. Starting at 1, what is the probability of hitting 3 before 0?
Answer: 4/7.
Explanation: Solve the harmonic recurrence or use [1 − (q/p)^i]/[1 − (q/p)^3] with i = 1. The boundary probabilities are 0 and 1.
A simple symmetric walk starts at 0 and is at 0 after 4 steps. What is the probability it was at 2 after 2 steps?
Answer: 1/6.
Explanation: Six equally likely step sequences return to zero. Only UUDD passes through 2 halfway; unconditional intermediate probabilities are inappropriate.
Each question has its own page with a short answer and a credit to its source. Hints and full worked solutions are in the question bank with a free account.
Our bank summary dated contains 928 Probability entries in the bank. Counts describe entries in our curated practice bank, including its source and classification choices. They do not measure employer interview frequency, candidate success rates or how often any one question is asked.
The public examples above were written for these guides. They illustrate the techniques and are not a reproduction or ranking of the bank's sourced questions. The dashboard's broader bank combines sourced questions, worked answers where available and references for book entries without a reproduced solution.
| Subtopic | Bank entries |
|---|---|
| Combinatorics | 188 |
| Expected Value | 177 |
| Conditional Probability | 131 |
| Events | 117 |
| Continuous Random Variables | 102 |
| Games & Strategy | 47 |
| Conditional Expectation | 41 |
| Discrete Random Variables | 33 |
| Covariance & Correlation | 22 |
| Random Walks | 18 |
| Markov Chains | 12 |
| Limit Theorems | 11 |
| Poisson Processes | 10 |
| Normal Distribution | 5 |
| Order Statistics | 5 |
| Geometric Probability | 4 |
| Martingales | 3 |
| Options Pricing | 1 |
| Algebra & Number Theory | 1 |
Bank difficulty labels: 448 easy · 356 medium · 124 hard.
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