This Exam P sample reference tests Combinatorial Probability. After one designated person is seated, the other has five available seats and exactly two are adjacent to the first. The probability is therefore 2/5, which is choice D.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis counts one favorable position out of all six seats, 1/6, even though one seat is already occupied and there are two neighboring positions.
BThis counts only one of the two seats neighboring the first designated person, giving 1/5.
CThis uses the line-arrangement adjacency probability 2/6=1/3 and fails to identify the two ends as adjacent in a circle.
EThis is the complementary probability 1-2/5=3/5 that at least one person separates the two designated people.
Original practice · fully worked
Original variant: separated trainees at a briefing
Eight distinct delegates sit uniformly at random around a circular briefing table. Three of the delegates are trainees. Calculate the probability that no two trainees occupy adjacent seats.
A 1/7
B 2/7
C 5/14
D 9/14
E 5/7
Variant answer in brief
Among C(8,3)=56 trainee seat sets on the circle, 16 contain no adjacent pair. The probability is 16/56=2/7, so choice B is correct.
Setup
Setup
Focus only on which three of the eight labeled circular seats are occupied by trainees.
Nall=(38)=56
Model
Model
Arrange the five nontrainee seats around the circle. They create five gaps, and separation requires placing the three trainees in three distinct gaps.
Ngap choices=(35)=10
Compute
Compute
For distinguished people, the gap construction gives the same probability as the full circular-arrangement count.
Pr(no adjacent trainees)=(8−1)!(5−1)!(35)3!
Pr(no adjacent trainees)=50401440=72
Answer
Answer
The probability that all three trainees are separated is 2/7.
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.