25 original Bayes and conditional-probability questions with answers, informed by conditional-probability coverage in our probability bank.
The observation protocol is often the entire problem. Learning that a named die is six differs from learning that at least one die is six; seeing two independent measurements differs from seeing one alert copied twice. This collection uses the conditional-probability coverage of our bank to organise 25 original examples around evidence, sampling and dependence. Read the condition aloud before selecting a denominator.
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.
Use prior-weighted likelihoods and make conditional independence explicit.
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.
Count the remaining ordered outcomes or update the composition of the sample.
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.
Observations can be independent given a hidden state while becoming dependent after that state is averaged out.
Use the previous two bags, replacing the first red token before another draw from the same chosen bag. Find the probability the next token is red.
Answer: 5/8.
Explanation: Mix the bag-specific red probabilities using the updated bag weights: (3/4)(3/4) + (1/4)(1/4).
Choose uniformly between a fair coin and one with heads probability 0.8. Observe one head. What is the probability the biased coin was chosen?
Answer: 8/13.
Explanation: Heads is more likely under the biased coin. Divide its prior-weighted likelihood 0.4 by total heads probability 0.65.
Choose uniformly between a fair coin and an 80%-heads coin. Toss that coin independently twice and see HH. Find probability it is biased.
Answer: 64/89.
Explanation: Conditional likelihoods are 0.25 and 0.64. The tosses share the hidden coin, so they are not marginally independent.
P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2. Are A and B independent?
Answer: Yes.
Explanation: The intersection equals P(A)P(B). Equivalently, conditioning on B leaves probability of A unchanged at 0.4.
A and B are disjoint and each has positive probability. Can they be independent?
Answer: No.
Explanation: Their intersection has probability zero, while the product of their positive probabilities is nonzero. Disjointness describes exclusion, not independence.
Renormalise densities or use a regular conditional distribution rather than dividing by a zero-probability event.
Flip two independent fair coins. Given that exactly one is heads, are their heads indicators independent?
Answer: No; they are perfectly negatively related.
Explanation: Only HT and TH remain. Once the first coin is known under this condition, the second is determined.
X is uniform on [0,10]. Given X > 4, what is P(X > 7)?
Answer: 1/2.
Explanation: The remaining interval has length 6 and the favourable interval length 3. Renormalise after conditioning.
X and Y are independent uniform [0,1]. Given X + Y > 1, what is P(X > 0.5)?
Answer: 3/4.
Explanation: The conditioning triangle has area 1/2. Its portion with X > 0.5 has area 3/8, giving ratio 3/4.
X and Y are jointly normal with zero means, unit variances and correlation 0.6. What is E[Y | X = 2]?
Answer: 1.2.
Explanation: The conditional Gaussian mean is correlation times X in this standardised case. The statement uses a regular conditional distribution, not division by P(X = 2).
T is exponential with rate 2 per minute. Given T > 1 minute, what is P(T > 2 minutes)?
Answer: exp(−2).
Explanation: Exponential memorylessness makes the remaining one-minute survival probability equal to survival from a fresh start.
The selected population may have different base rates. Reverse a conditional only with the appropriate marginal probabilities.
A homogeneous Poisson process has 6 arrivals in an hour. What is the probability exactly 2 occurred in the first half-hour?
Answer: 15/64.
Explanation: Conditional on the total, the half-hour count is binomial with n = 6 and p = 1/2. The original arrival rate cancels.
Half the observations are from group A and half from B. Selection retains 80% of A and 20% of B. What fraction of retained observations are A?
Answer: 80%.
Explanation: Among one unit of original population, retained masses are 0.4 and 0.1. Conditioning on selection changes group proportions.
A market is in regime A with probability 0.25. A signal fires with probability 0.6 in A and 0.2 otherwise. Find its firing probability.
Answer: 0.3.
Explanation: Average conditional signal probabilities over an exhaustive, disjoint partition: 0.25 × 0.6 + 0.75 × 0.2.
If X is continuous, can you calculate P(Y > 0 | X = x) using P(Y > 0, X = x)/P(X = x)?
Answer: No; the ratio is 0/0.
Explanation: Use a conditional density or regular conditional distribution with appropriate assumptions. The elementary event formula requires a positive-probability condition.
If P(A | B) = 0.9, must P(B | A) equal 0.9?
Answer: No.
Explanation: Reversing the conditional requires the base rates through Bayes. A high likelihood alone does not determine the posterior probability.
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