This Exam P sample reference tests Expected Value. Every covered employee selects exactly two items, so the sum of the three marginals is twice the selection probability. The remaining probability is 1/2, choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value zero results from treating the marginal sum 1 as the probability that some pair is selected and taking its complement. Each selected pair is counted twice in that sum.
DThe value 97/144 is the complement of choice B, but neither fraction follows from the required relation 2q=1.
Original practice · fully worked
Original variant: none selected with single-or-triple activation
Users either activate no dashboard widget, exactly one of widgets A, B, and C, or all three. The three activation marginals are 0.25, 0.30, and 0.35, and 0.08 of users activate all three. Find the probability that a user activates no widget.
A 0.26
B 0.18
C 0.24
D 0.66
E 0.74
Variant answer in brief
The marginal sum counts each single activation once and each triple activation three times. Thus exactly-one mass is 0.66, total active mass is 0.74, and none is 0.26, choice A.
Setup
Setup
Let N be the number of activated widgets and add the three widget marginals to obtain the expected number activated.
m=0.25+0.30+0.35=0.90
Model
Model
A one-widget user contributes one to the marginal sum and a three-widget user contributes three, so solve for the exactly-one mass.
m=Pr(N=1)+3Pr(N=3),Pr(N=1)=0.90−3(0.08)=0.66
Compute
Compute
The exactly-one mass is 0.90-3(0.08)=0.66. Subtracting the exactly-one and exactly-three masses from one leaves 0.26 for no widget.
Pr(N=0)=1−0.66−0.08=0.26
Answer
Answer
Thus the no-widget probability is 0.26, corresponding to choice A.
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