This Exam P sample reference tests Inclusion–Exclusion. Summing the three risk-factor memberships counts single-factor participants once, double-factor participants twice, and triple-factor participants three times. This identifies 174 people with exactly two factors. Adding every disjoint region, including the 155 with none, gives 617 and choice C.
Let N₂ be the number of participants with exactly two risk factors. Total the three single-factor-only regions and the three-factor region.
N1=3(68)=204,N3=84
Model
Model
Count risk-factor memberships in two ways. The three reported factor totals count each exactly-two participant twice and each exactly-three participant three times.
3(268)=N1+2N2+3N3
Compute
Compute
Solve for the exactly-two group, then add the disjoint participant groups.
804=204+2N2+252,N2=174
Ntotal=204+174+84+155=617
Answer
Answer
The study requires 617 people in total.
617(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 443 adds the single-factor-only, all-three, and no-factor groups but omits all 174 participants with exactly two factors.
BThe value 462 is the number of participants with at least one risk factor. It omits the 155 participants with none.
DA total of 636 would place 193 participants in the exactly-two regions. Their membership count would make the three factor totals sum to 842 rather than 804.
EThe value 791 adds the 174 exactly-two participants a second time. Those people have already been included once as individuals in the total 617.
Original practice · fully worked
Original variant: junctions flagged by overlapping detectors
A utility inspects its junctions with thermal, acoustic, and vibration detectors. The detectors flag 140, 120, and 100 junctions, respectively. The three pairwise intersection counts are 40, 30, and 20, and 10 junctions are flagged by all three detectors. Calculate the number of junctions flagged by at least one detector.
A 260
B 270
C 280
D 290
E 360
Variant answer in brief
Inclusion–exclusion adds the three detector counts, subtracts the three pairwise intersections, and restores the all-three intersection once. This gives 280 junctions and choice C.
Setup
Setup
Let T, A, and V denote the three detector flag sets.
∣T∣+∣A∣+∣V∣=140+120+100=360
Model
Model
Apply inclusion–exclusion to the union of the three flag sets.
∣T∪A∪V∣=∣T∣+∣A∣+∣V∣
−∣T∩A∣−∣T∩V∣−∣A∩V∣+∣T∩A∩V∣
Compute
Compute
Substitute the three pairwise counts and the all-three count.
∣T∪A∪V∣=360−(40+30+20)+10
=280
Answer
Answer
Exactly 280 junctions are flagged by at least one detector.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
