This Exam P sample reference tests Joint Cumulative Distribution Functions. This joint-CDF event combines a count restriction with an aggregate-loss restriction. Counts zero and one always qualify, while a count of two qualifies unless both severities equal two, giving 0.9575 and choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.0250 is P(X=2,Y=3): two events occur and exactly one of their severities equals two. It omits all other outcomes in the cumulative event.
BThe value 0.0500 is P(X=2) and ignores both the aggregate restriction and the qualifying counts zero and one.
CThe value 0.0675 is the complement of 0.9325. It results from first evaluating the wrong event Y≤2 and then taking its complement.
DThe value 0.9325 equals 0.80+0.12+0.05(0.25), which imposes Y≤2 rather than Y≤3 when two events occur.
Original practice · fully worked
Original variant: nonuniform joint calibration scores
Normalized calibration scores U and V have joint density f(u,v)=(6/5)(u+v²) for 0≤u≤1 and 0≤v≤1, and zero elsewhere. Let G be their joint cumulative distribution function. Calculate G(0.80,0.50).
A 0.040
B 0.192
C 0.232
D 0.400
E 0.704
Variant answer in brief
Integrate the nonconstant joint density over the CDF rectangle. The u and v-squared terms contribute 0.192 and 0.040, totaling 0.232 and selecting choice C.
Setup
Setup
The required event is the rectangle from the origin to u=0.80 and v=0.50. The scale factor normalizes the density over the full unit square.
∫01∫0156(u+v2)dvdu=56(21+31)=1
G(0.80,0.50)=Pr(U≤0.80,V≤0.50)
Model
Model
Integrate the density over the bounded rectangle and keep its two additive contributions separate.
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