Original practice · fully worked
Original variant: two relative-order requirements
A service desk receives seven distinct repair tickets labeled A through G and schedules them in a uniformly random order. Calculate the probability that ticket A is scheduled earlier than both B and C, while ticket D is scheduled later than E.
- A
0.0714 - B
0.1667 - C
0.3333 - D
0.5000 - E
0.8333
Answer and calculation
Among the relative orders of A, B, and C, ticket A is earliest in two of the six orders, giving probability 1/3. Independently, D is later than E with probability 1/2. Their product is 1/6, approximately 0.1667 and choice B.
Setup
Setup
Focus on relative order within the two disjoint ticket groups; the positions of the remaining tickets do not affect either requirement.
{A,B,C},{D,E} Model
Model
Each member of the three-ticket group is equally likely to be earliest, and either order of D and E is equally likely.
Pr(A earliest among A,B,C)=31 Pr(D later than E)=21 Compute
Compute
Relative orders of disjoint labeled groups are independent, so multiply the two symmetry probabilities.
Pr(both requirements)=31⋅21=61=0.1666667… Answer
Answer
Both relative-order conditions hold with probability about 0.1667.
0.1667(B)