25 challenging original probability questions with answers on patterns, barrier hitting, hidden parameters, renewal sampling and martingale conditions.
Our bank labels a substantial subset of probability questions Hard. This editorial collection targets the reasons a problem becomes challenging: overlapping states, length-biased sampling, latent parameters and unjustified stopping arguments. It is not a measured ranking of the hardest questions or a claim about candidate failure rates. Attempt the setup before reading the explanation, then use the boundary or integrability check to test your own derivation.
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.
Track just enough history for a valid recurrence. Endpoint conditioning changes which paths remain possible.
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.
The generator supplies a boundary-value problem; time-based sampling reweights intervals by their lengths.
Standard Brownian motion starts at x in (0,a). What is the probability it hits a before 0?
Answer: x/a.
Explanation: Stop the bounded process on exiting the interval and use its martingale mean. The bounded stopped process justifies the stopping argument.
Standard Brownian motion starts at x in (0,a). Find expected time to hit either boundary.
Answer: x(a − x).
Explanation: The mean exit time solves u″/2 = −1 with zero boundary values. The generator includes the factor 1/2.
For standard Brownian motion started at zero and a > 0, find P(maximum by time T ≥ a).
Answer: 2[1 − Φ(a/√T)].
Explanation: Reflect paths after their first crossing of a. The factor two relates a pathwise maximum to a terminal normal tail.
Every renewal interval lasts exactly 1 or 3 minutes with equal probability. At a stationary random time, what is expected length of the interval containing you?
Answer: 2.5 minutes.
Explanation: Time sampling weights intervals by length, giving probabilities 1/4 and 3/4. Thus the mean is E[L²]/E[L] = 5/2.
In the same stationary renewal process with interval lengths 1 or 3 equally likely, find expected remaining wait.
Answer: 1.25 minutes.
Explanation: Within a time-sampled interval, the location is uniform. The residual mean is E[L²]/(2E[L]) = 5/4.
Exchangeable spacings and posterior updates can replace lengthy integrations when the assumptions fit.
Sort two independent uniform [0,1] points. What is the probability their gap exceeds 1/2?
Answer: 1/4.
Explanation: In the unit square, |X − Y| > 1/2 occupies two triangles, each area 1/8. Ordering the points does not create a new sampling rule.
Four observations are independent uniform [0,1]. What is the expected second-smallest observation?
Answer: 2/5.
Explanation: The five spacings are exchangeable and sum to 1. The second order statistic is the sum of the first two spacings.
A coin's heads probability p is uniform on [0,1]. Conditional on p, tosses are independent. Find P(second head | first head).
Answer: 2/3.
Explanation: One head changes the posterior density to 2p. Its mean is 2/3; shared parameter uncertainty creates marginal dependence.
A coin has uniform prior for p. After 3 heads and 1 tail, find the predictive probability of the next head.
Answer: 2/3.
Explanation: The posterior is Beta(4,2), whose mean is 4/6. The prior contributes one pseudo-count to each outcome.
A Poisson process has rate 5 per hour. Independently mark each arrival red with probability 0.4. Are red and non-red counts in a fixed hour independent?
Answer: Yes; they are Poisson with means 2 and 3.
Explanation: Independent thinning splits a Poisson process into independent processes. Conditioning on their total would instead make the counts dependent.
Check whether a seemingly symmetric answer has a finite expectation and whether a stopping time supports the claimed identity.
Independent processes A and B have Poisson rates 2 and 3. What is the probability A's second arrival occurs before B's first?
Answer: 4/25.
Explanation: In the merged process, each arrival is A with probability 2/5 independently. The first two must both be A.
N is Poisson with mean 3. Independent severities have mean 2 and variance 5, independent of N. Find variance of their total.
Answer: 27.
Explanation: Total variance gives E[N]Var(X) + Var(N)E[X]² = 3 × 5 + 3 × 4.
X and Y are jointly standard normal with correlation 1/2. Find P(X and Y have opposite signs).
Answer: 1/3.
Explanation: The Gaussian quadrant identity gives P(X > 0,Y > 0) = 1/4 + arcsin(ρ)/(2π). Subtract both same-sign quadrants from 1.
X and Y are independent standard normal variables. Does E[X/Y] exist as an ordinary expectation?
Answer: No.
Explanation: The ratio is Cauchy. Its positive and negative parts both have infinite expectation, so symmetry does not produce a defined mean.
A fair integer random walk starts at 0 and stops when it first reaches 1. Is its expected stopping time finite?
Answer: No, although it reaches 1 almost surely.
Explanation: Almost-sure finiteness does not imply finite expectation. An unbounded stopping argument cannot simply transfer the martingale mean to this time.
Distinguish a sharp calculation from a bound, and select the correct fixed point or event sequence.
M is a continuous square-integrable martingale with M₀ = 0 and E[M_T²] = 9. Give a general upper bound on E[supₜ≤T |Mₜ|²].
Answer: 36.
Explanation: Doob's L² maximal inequality bounds this expectation by four times the terminal second moment. It is an upper bound, not an equality.
Each individual has zero children with probability 0.4 and two with probability 0.6, independently. What is the extinction probability from one ancestor?
Answer: 2/3.
Explanation: The extinction probability is the smallest root in [0,1] of q = 0.4 + 0.6q². The other root, 1, is not the extinction probability here.
A coin has uniform prior for p. After 3 heads and 1 tail, let X and Y indicate heads on two future conditionally independent tosses. Find their posterior predictive covariance.
Answer: 2/63.
Explanation: The posterior is Beta(4,2). Conditional covariance is zero, so total covariance equals Var(p | data) = 4 × 2/(6² × 7). Shared uncertainty couples the future tosses.
X, Y, Z are independent exponential variables with the same rate. Find P(X > Y + Z).
Answer: 1/4.
Explanation: Condition on Y + Z. The survival of X gives E[exp(−λY)]E[exp(−λZ)] = (1/2)².
Three independent exponential clocks have common rate λ. Find the expected last ring time.
Answer: 11/(6λ).
Explanation: Successive gaps have rates 3λ, 2λ and λ by memorylessness. Add their means, rather than multiplying the largest mean by three.
Our bank summary dated contains 124 probability entries labelled Hard 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.
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