This Exam P sample reference tests Joint Distributions. Grouping the nine joint outcomes by whether their maximum is zero, one, or two gives masses 0.38, 0.37, and 0.25. The resulting expected maximum is 0.87, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis computes the expected minimum. The positive-minimum cells contribute 0.11+0.03+0.09+2(0.05)=0.33.
BThis reports E[Y]=1(0.30)+2(0.20)=0.70, using only one marginal rather than the maximum of both variables.
DThis reports E[X+Y]=E[X]+E[Y]=0.50+0.70=1.20. A maximum is not the sum.
EThis reports the largest possible table value, 2, without probability-weighting the joint outcomes.
Original practice · fully worked
Original variant: expected peak diagnostic severity
A diagnostic system records a daily peak severity M in {0,1,2,3}. Historical tail probabilities are P(M≥1)=0.72, P(M≥2)=0.35, and P(M≥3)=0.10. Calculate E[M].
A 0.10
B 0.35
C 0.72
D 1.17
E 1.72
Variant answer in brief
A bounded nonnegative integer variable has mean equal to the sum of its positive tail probabilities. Adding the three supplied tails gives 1.17, so choice D is correct.
Setup
Setup
Represent the severity as the number of positive thresholds it crosses.
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