This Exam P sample reference tests Bernoulli Distribution. For Bernoulli indicators, E[XY] is the probability that both indicators equal one. Solving the correlation identity gives E[XY]=0.005+0.30sqrt(0.10·0.90·0.05·0.95)=0.024615, which rounds to 0.025 and selects choice C.
Represent the two one-year claim counts by Bernoulli indicators X and Y. Their means and standard deviations follow from the two marginal probabilities.
E[X]=0.10,SD(X)=0.10(0.90)=0.30
E[Y]=0.05,SD(Y)=0.05(0.95)
Model
Model
Insert the unknown joint moment into the definition of correlation. Because X and Y are indicators, XY equals one exactly when both claims occur.
ρ=SD(X)SD(Y)E[XY]−E[X]E[Y]
E[XY]=Pr(X=1,Y=1)
Compute
Compute
Rearrange the correlation equation and substitute the two Bernoulli variances.
Pr(X=1,Y=1)=0.10(0.05)+0.300.10(0.90)0.05(0.95)
Pr(X=1,Y=1)=0.0246150452…
Answer
Answer
The joint claim probability rounds to 0.025.
0.025(C)
Continue without hunting through PDFs
All 718 Exam P sample solutions in syllabus order
The searchable 3108-page manual includes this complete solution, its error analysis, and one original worked variant for every active reference.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis multiplies the marginals, 0.10(0.05)=0.005, as though the indicators were independent. Their positive correlation makes the joint probability larger.
BA joint probability of 0.006 would produce correlation (0.006-0.005)/√(0.10(0.90)0.05(0.95))=0.0153, not the supplied 0.30.
DA joint probability of 0.033 would imply correlation about 0.428. This results from adding too much covariance to the independent baseline.
EThe value 0.045 equals 0.05-0.005, the Y-only cell under an independence calculation. It is neither the both-claim cell nor compatible with correlation 0.30.
Original practice · fully worked
Original variant: recover alert correlation from a union
During a production cycle, indicator A records a temperature alert and indicator B records a vibration alert. Their probabilities are P(A=1)=0.40 and P(B=1)=0.50, while the probability of at least one alert is 0.65. Calculate the correlation between A and B.
A -0.2041
B 0.0000
C 0.0500
D 0.2041
E 0.2500
Variant answer in brief
Inclusion-exclusion gives P(A=1,B=1)=0.40+0.50-0.65=0.25. The covariance is 0.25-0.20=0.05, and standardizing it gives 0.05/√(0.24·0.25)=0.2041, so choice D.
Setup
Setup
Use the union probability to recover the one-one cell of the two-indicator table.
Pr(A=1,B=1)=0.40+0.50−0.65=0.25
Model
Model
Compute the covariance and both Bernoulli variances from the recovered joint and the supplied marginals.
Cov(A,B)=0.25−(0.40)(0.50)=0.05
Var(A)=0.40(0.60)=0.24,Var(B)=0.50(0.50)=0.25
Compute
Compute
Standardize the covariance by the product of the two standard deviations.
Corr(A,B)=(0.24)(0.25)0.05
Corr(A,B)=0.2041241452…
Answer
Answer
The alert indicators have correlation approximately 0.2041.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
