This Exam P sample reference tests Set Probability. The three single-factor-only regions contain 1,260 people. Symmetry forces each pair-only region to contain 230, so the three pair-only regions contain 690; adding 320 in all three and 480 in none gives 2,750 and choice C.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis adds only the exactly-one and no-factor groups, 1260+480=1740, omitting everyone with two or three factors.
BThis adds only one of the three 230-person pair-only regions: 1260+230+320+480=2290.
DThe aggregate two-factor membership count is 1380, but every exactly-two person contributes twice. Using 1380 as a person count gives 1260+1380+320+480=3440.
EThis adds the three set sizes and the no-factor group, 3(1200)+480=4080, without removing people counted in multiple sets.
Original practice · fully worked
Original variant: servers with exactly one alert class
Among 1,000 servers, alert A appears on 420, alert B on 380, and alert C on 350. The pairwise intersections contain 160, 140, and 120 servers for A and B, A and C, and B and C, respectively. Exactly 800 servers have at least one alert. Calculate the number of servers with exactly one alert class.
A 70
B 200
C 420
D 520
E 730
Variant answer in brief
Inclusion-exclusion first gives 70 servers in all three alert classes. The three exactly-one regions are then 190, 170, and 160, totaling 520 and selecting choice D.
Setup
Setup
Use the known union to solve for the triple intersection.
800=(420+380+350)−(160+140+120)+NABC
Model
Model
Recover the all-three count, then remove both pairwise overlaps from each marginal and restore the triple intersection once.
NABC=70
NA-only=420−160−140+70=190
NB-only=380−160−120+70=170
NC-only=350−140−120+70=160
Compute
Compute
Add the three disjoint exactly-one regions.
Nexactlyone=190+170+160=520
Answer
Answer
There are 520 servers with exactly one alert class.
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