Worked solution · Exam P
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.
Independent solution
Setup
Use rotational symmetry to place one designated person at a fixed seat.
Model
On a circle, the fixed seat has two distinct neighboring seats.
Compute
The second designated person is equally likely to occupy any of the five remaining seats.
Answer
The two designated people are adjacent with probability 2/5.
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Original practice · fully worked
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.
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
Focus only on which three of the eight labeled circular seats are occupied by trainees.
Model
Arrange the five nontrainee seats around the circle. They create five gaps, and separation requires placing the three trainees in three distinct gaps.
Compute
For distinguished people, the gap construction gives the same probability as the full circular-arrangement count.
Answer
The probability that all three trainees are separated is 2/7.
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