This Exam P sample reference tests Mutually Exclusive Events. A union probability equals the sum of three event probabilities only when the selected events are pairwise disjoint. A, B, and E cannot occur in any pair, whereas every other listed triple contains an overlapping pair, so choice A is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
BIn the triple A, C, E, events C and E can occur together: one first-year theft, none in year two, and at least one in year three. Their probabilities therefore cannot simply be added.
CIn the triple A, D, E, event A is contained in D because no thefts over all three years implies none in year three. This creates a nonempty overlap.
DEvents B, C, and D can all occur together: at least one second-year theft, exactly one first-year theft, and no third-year theft are compatible.
EAlthough B and E are disjoint, C can occur with either B or E. Pairwise disjointness fails for the triple B, C, E.
Original practice · fully worked
Original variant: quantify the failure of event additivity
A fair six-sided selector outputs one of the integers 1 through 6. Define U={1,2,3}, V={3,4}, and W={4,5}. Calculate [P(U)+P(V)+P(W)]-P(U union V union W).
A 1/12
B 1/3
C 2/3
D 5/6
E 7/6
Variant answer in brief
The three individual probabilities sum to 7/6, while their union contains five of the six outputs and has probability 5/6. The additivity gap is 2/6=1/3, choice B.
Setup
Setup
Compute the three event probabilities from their numbers of selector outcomes.
Pr(U)=63,Pr(V)=62,Pr(W)=62
Model
Model
Identify the distinct outputs included in at least one event.
U∪V∪W={1,2,3,4,5}
Pr(U∪V∪W)=65
Compute
Compute
Subtract the union probability from the sum of the marginals.
(63+62+62)−65=62=31
Answer
Answer
The overlap creates an additivity gap of one third.
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