Given P(A and B)=0.12 and P(B)=0.4, find P(A∣B).
Answer and explanation
0.3
Divide 0.12 by 0.4.
Subjects
Probability models
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Sample exercises
X takes values 0,2,5 with probabilities 0.2,0.5,0.3. Find E[X].
Answer and explanation
2.5
Compute 0(0.2)+2(0.5)+5(0.3)=2.5.
Two events have positive probability and are disjoint. Are they independent?
- Always
- Only if their names differ
- No
- Independence is undefined for events
Answer and explanation
No
Their intersection probability is zero while the product of their positive probabilities is positive.
Independent X and Y have variances 4 and 9. Find Var(2X−Y).
Answer and explanation
25
Variance scales by the square of coefficients, giving 4×4+9=25.
Show by an example that E[XY] need not equal E[X]E[Y].
Answer and explanation
Let X=Y be a fair 0/1 variable. Then E[XY]=E[X]=1/2, while E[X]E[Y]=1/4.
Credit a dependent example and both computed quantities.
Does E[X+Y]=E[X]+E[Y] require independence?
- Yes in every case
- Only if the means are zero
- Only for normal distributions
- No, assuming expectations exist
Answer and explanation
No, assuming expectations exist
Linearity of expectation does not require independence.
Study notes
Conditional probability and expectation
P(A∣B)=P(A and B)/P(B) requires P(B)>0.
UniversityConditional probability and expectation · revision notes
P(A∣B)=P(A and B)/P(B) requires P(B)>0.
UniversityContinuous distributions and densities
A density is integrated over a set; its value at a point is not the point probability.
UniversityContinuous distributions and densities · revision notes
A density is integrated over a set; its value at a point is not the point probability.
UniversityProbability rules
A possible result of a random process.
UniversityProbability rules · revision notes
A possible result of a random process.
UniversityVariance of a binomial count
Assume a fixed number of independent trials with the same success probability.
UniversityVariance of a binomial count · formula reference
Assume a fixed number of independent trials with the same success probability.
UniversityBayes rule for an abstract classifier
Use the base rate and both conditional likelihoods. This is an abstract classification model, not a medical test.
UniversityBayes rule for an abstract classifier · formula reference
Use the base rate and both conditional likelihoods. This is an abstract classification model, not a medical test.
UniversityPoisson zero-event probability
λ is the mean count for the entire specified interval, not a rate without a time unit.
UniversityPoisson zero-event probability · formula reference
λ is the mean count for the entire specified interval, not a rate without a time unit.
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