This Exam P sample reference tests Joint Distributions. This problem recovers two adjacent point masses from a joint cumulative table. A four-corner rectangle difference gives 0.05, which selects choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe mismatched-corner calculation |0.84-0.76-0.62+0.53|=0.01 pairs inconsistent x-boundaries. Rectangle subtraction must use the same two x-boundaries at both y-boundaries.
BThis is p(4,9)=0.76-0.65-0.62+0.53=0.02, so it omits the mass at x=5.
CThis is p(5,9)=0.84-0.76-0.67+0.62=0.03, so it omits the mass at x=4.
DThis doubles the 0.02 mass at x=4, incorrectly assuming the two adjacent point masses are equal.
Original practice · fully worked
Original variant: selected jumps of a cumulative count distribution
A warehouse records the integer number N of delayed parcels in a dispatch. Its cumulative probabilities are F(1)=0.18, F(2)=0.41, F(3)=0.68, and F(4)=0.86. Calculate the probability that a dispatch has exactly two or exactly four delayed parcels.
A 0.18
B 0.23
C 0.41
D 0.45
E 0.68
Variant answer in brief
Each point probability is a jump of the discrete CDF. The jumps at 2 and 4 are 0.23 and 0.18, so the requested probability is 0.41 and choice C.
Setup
Setup
For an integer-valued variable, isolate a point mass by subtracting consecutive CDF values.
Pr(N=k)=F(k)−F(k−1)
Model
Model
The two requested counts are disjoint, so their masses add.
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