Original practice · fully worked
Original variant: exactly one detector flags a junction
Three monitoring teams review the same set of road junctions. Historical proportions flagged by teams A, B, and C are 0.35, 0.25, and 0.20. Pairwise overlap proportions are 0.10, 0.08, and 0.05, and all three teams flag 0.02. Find the proportion flagged by exactly one team.
- A
0.30 - B
0.34 - C
0.40 - D
0.46 - E
0.60
Answer and calculation
Exactly-one membership equals the marginal sum minus twice the pairwise-overlap sum plus three times the triple overlap, giving 0.40, choice C.
Setup
Setup
Let p-one denote the proportion of junctions flagged by exactly one of the three teams.
p1=Pr(exactly one flag) Model
Model
In the marginal sum, an exactly-two junction is counted twice and an exactly-three junction three times. Subtracting twice each pairwise overlap and restoring three times the triple overlap leaves exactly-one membership.
p1=∑Pr(Ai)−2∑Pr(Ai∩Aj)+3Pr(A∩B∩C) Compute
Compute
The marginal probabilities sum to 0.80, the pairwise overlaps sum to 0.23, and the triple overlap is 0.02; substitution gives 0.40.
p1=0.80−2(0.23)+3(0.02)=0.40 Answer
Answer
Therefore 40% of junctions are flagged by exactly one team, which is choice C.
0.40(C) 