This Exam P sample reference tests Set Probability. Because B is contained in A while C is disjoint from both, the three union probabilities are 0.30, 1.00, and 0.80. Their sum is 2.10, so choice C is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis adds only the three marginal probabilities, 0.30+0.10+0.70=1.10, rather than evaluating each union.
BThis applies an independence product correction even though the event definitions give containment and disjointness.
DThis respects disjointness with C but treats A and B as independent, using 0.30+0.10-0.03=0.37 for their union and obtaining 0.37+1.00+0.80=2.17.
EThis treats A and B as disjoint in A union B, using 0.40 instead of 0.30 and raising the total to 2.20.
Original practice · fully worked
Original variant: nested quality-control flags
In a production audit, R is the event that an item receives a red flag, Q is the event that it receives any quality flag, and S is the event that it is diverted for a separate random inspection. Every red flag is a quality flag, and diversion is disjoint from either flag event. Suppose P(Q)=0.42, P(R)=0.18, and P(S)=0.31. Calculate P(Q union R)+P(Q union S)+P(R union S).
A 0.91
B 1.40
C 1.46
D 1.64
E 1.82
Variant answer in brief
R is contained in Q, while S is disjoint from both. The three unions are 0.42, 0.73, and 0.49, which sum to 1.64. Choice D is correct.
Setup
Setup
Translate the stated relationships into set notation.
R⊂Q,Q∩S=∅
Model
Model
Containment fixes the first union, while disjointness turns the other two into sums.
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