40 original expected-value questions with answers, grouped by technique and informed by expected-value coverage in our probability bank.
Expected value is one of the largest labelled subtopics in our probability question bank. That makes it a useful place to build technique coverage: indicator sums, geometric waits, recursion, total expectation and optional stopping all turn up in different guises. These 40 original examples are grouped by method, rather than presented as an employer-frequency ranking. Each includes its answer and the key step that makes the calculation work.
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.
For counts, add indicator expectations. For a trading payoff, separate proceeds from costs and losses.
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.
A geometric wait comes from a fresh independent attempt; maxima and minima often yield to survival functions.
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.
Dependence does not prevent an indicator calculation, but overlapping waiting patterns may require multiple states.
A nonnegative integer N satisfies P(N ≥ k) = 2^(−k) for k ≥ 1. Find E[N].
Answer: 1.
Explanation: Add the survival probabilities over positive integers. The geometric series starts at 1/2, not at 1.
Six orders independently choose one of three venues uniformly. What is the expected number of venues receiving an order?
Answer: 3[1 − (2/3)^6] = 665/243.
Explanation: For each venue, use an indicator for at least one arrival. Its complement is that all six orders chose elsewhere.
A uniformly random permutation matches 10 labels to 10 records. How many matches are correct on average?
Answer: 1.
Explanation: Every label has a 1/10 chance of its original position. The indicators are dependent, but their expectations still add.
In 8 independent fair coin flips, how many adjacent HH pairs occur on average, allowing overlaps?
Answer: 7/4.
Explanation: There are seven adjacent positions, each HH with probability 1/4. Overlapping pairs do not invalidate the expectation calculation.
With independent fair coin flips, what is the expected wait through the first HH?
Answer: 6 flips.
Explanation: Use states with zero or one trailing H. E₀ = 1 + E₀/2 + E₁/2 and E₁ = 1 + E₀/2.
Use stages for coupon collection and total expectation when a hidden regime or sampling rule changes the distribution.
Each independent draw gives one of four equally likely labels. Find the expected draws to see every label.
Answer: 25/3 draws.
Explanation: Once k labels are seen, a new label arrives with probability (4 − k)/4. Add waiting means 1 + 4/3 + 2 + 4.
A book contains 7 marked and 13 unmarked records. Sample 5 uniformly without replacement. Find expected marked records.
Answer: 7/4 records.
Explanation: Each sampled position has marked probability 7/20. Dependence between draws affects variance, not the sum of their means.
Randomly order 3 marked and 9 unmarked records. What is the expected position of the first marked record?
Answer: 13/4.
Explanation: The 9 unmarked records occupy four exchangeable gaps around marked records. The first gap averages 9/4, then add one.
A desk is busy with probability 0.3. Expected arrivals are 20 when busy and 4 when quiet. What is unconditional expected arrivals?
Answer: 8.8.
Explanation: Apply total expectation: 0.3 × 20 + 0.7 × 4. Regime means must be weighted by regime probabilities.
A random variable has conditional means 2 and 6 in equally likely regimes, and conditional variance 1 in each. Find its variance.
Answer: 5.
Explanation: Total variance is expected conditional variance plus variance of conditional means: 1 + 4. Averaging within-regime variances misses regime uncertainty.
Condition on the count before summing random costs, and distinguish arithmetic means from compounded growth.
N is Poisson with mean 4. Independent per-order costs have mean 2 and are independent of N. Find expected total cost.
Answer: 8 units.
Explanation: Condition on N: the expected sum is 2N. Taking another expectation yields 2E[N]. State the independence assumption.
Incidents arrive at rate 0.5 per hour for 8 hours. Each incident independently costs 12 units on average. Find expected total loss.
Answer: 48 units.
Explanation: Expected incident count is rate times time, or 4. Multiply by mean severity; this says nothing about the loss quantile.
Two independent exponential clocks have rates 2 and 3 per minute. What is expected time until the first ring?
Answer: 1/5 minute.
Explanation: Their minimum has survival exp(−5t), so it is exponential with the sum of the rates.
X is uniform on [0,10]. Given X > 6, what is E[X]?
Answer: 8.
Explanation: Conditioning leaves a uniform distribution on [6,10]. Its midpoint gives the mean; do not keep the unconditional midpoint 5.
A positive gross return R equals 0.5 or 1.5 with equal probability. Compare E[R] and exp(E[log R]).
Answer: E[R] = 1; geometric mean is √0.75, approximately 0.866.
Explanation: Concavity of log separates expected wealth from typical compounded growth. A fair arithmetic return can have negative expected log growth.
The sampling unit changes weights; an optional action must be compared with its conditional continuation value.
Equally many batches have sizes 1 and 9. Choose an order uniformly from all orders. What is its expected batch size?
Answer: 8.2 orders.
Explanation: The chosen order belongs to a large batch with probability 9/10. Weight sizes by order counts: 0.1 × 1 + 0.9 × 9.
Roll a fair die. You may keep the result or reroll once and must accept the new result. Find the optimal expected payoff.
Answer: 4.25.
Explanation: Reroll 1, 2, 3 because the fresh mean is 3.5. Average max(face, 3.5) across the initial roll.
Roll a fair die. One optional reroll costs 1 unit and its face value must be accepted. What initial scores should you reroll?
Answer: Reroll 1 and 2; expected net payoff is 23/6.
Explanation: The reroll has net value 3.5 − 1 = 2.5. Compare each observed score against that value before averaging.
Independent offers are uniform on [0,1]. Keep drawing until an offer exceeds 0.8, with no cost or limit. Find expected accepted value and number of draws.
Answer: Accepted value 0.9; expected draws 5.
Explanation: Accepted offers are uniform on [0.8,1]. The number of draws is geometric with success probability 0.2; these are different expectations.
A payment of 1 arrives after an exponential waiting time T with rate 2, measured in years. At continuous discount rate 1, find E[exp(−T)].
Answer: 2/3.
Explanation: Integrate exp(−t) against density 2exp(−2t). Discounting at the mean arrival time would give a different answer.
Write the region where the payoff is positive and impose boundary conditions on a waiting-time recurrence.
S is uniform on [0,2]. Find E[(S − 1)⁺]. Ignore discounting.
Answer: 1/4.
Explanation: Integrate (s − 1)/2 over s from 1 to 2. Half the outcomes pay zero; averaging only positive payoffs doubles the result.
Loss L is uniform on [0,100]. Find E[(L − 80)⁺].
Answer: 2 units.
Explanation: Integrate the excess loss over [80,100] against density 1/100. This is an unconditional excess, not the conditional tail mean.
For n independent uniform [0,1] variables, find the expected maximum.
Answer: n/(n + 1).
Explanation: Its CDF is xⁿ, so integrate 1 − xⁿ over [0,1]. The result approaches 1 as sample size grows.
Three observations are independent uniform [0,1]. What is the expected range?
Answer: 1/2.
Explanation: Expected maximum is 3/4 and expected minimum is 1/4. Subtract their means; no independence between extrema is needed.
A simple symmetric random walk starts at 2 and stops on hitting 0 or 5. Find expected number of steps.
Answer: 6.
Explanation: The mean exit time from state i solves eᵢ = 1 + (eᵢ₋₁ + eᵢ₊₁)/2 with zero boundary values. Its solution is i(5 − i).
Verify finiteness and exploit linearity, exchangeability or a tail integral before reaching for a full distribution.
Each individual independently has mean 0.8 children. Starting from one individual, what is expected total population across all generations, including the first?
Answer: 5.
Explanation: Generation k has mean 0.8^k. Add the convergent geometric series; the subcritical mean is essential for a finite total expectation.
X and Y have finite means 3 and 2. Find E[X − 1.5Y], regardless of their dependence.
Answer: 0.
Explanation: Linearity gives 3 − 1.5 × 2. Zero mean exposure does not imply zero variance or a riskless hedge.
X has mean 2 and variance 9. What is E[X²]?
Answer: 13.
Explanation: Rearrange Var(X) = E[X²] − E[X]². This is a frequent reason that replacing a random input by its mean fails.
X₁, X₂, X₃ are identically distributed independent positive variables. What is E[X₁/(X₁ + X₂ + X₃)]?
Answer: 1/3.
Explanation: The three shares sum to 1 and have equal expectations by exchangeability. Taking a ratio of expectations is unnecessary.
X is exponential with rate 2. Find E[min(X,1)].
Answer: (1 − exp(−2))/2.
Explanation: Integrate P(X > t) from 0 to the cap 1. Tail integration naturally includes the probability mass at the cap.
Our bank summary dated contains 177 Expected Value entries in the probability 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.
Use these lesson previews to find the definitions and derivations behind the worked examples.
Continue with 928 probability questions in the practice bank.
Open the probability bank