Question
A reception desk repeats the same schedule of four back-to-back service intervals: first 3 minutes, then 7, then 2, then 8. After the final interval, the 20-minute cycle immediately starts again. An inspector arrives at a time chosen uniformly from one complete cycle; arrival exactly at a boundary has probability zero.
What is the probability that more than four minutes remain in whichever interval contains the arrival? It is not correct to regard the four intervals as equally likely to be visited.
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Hint 1
A uniformly sampled moment is more likely to fall inside a long interval than a short one.
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Solution
The inspector selects a time, not one of four interval labels. In an interval of length \(L\), an arrival leaves more than four minutes remaining exactly when it occurs in the first \(L-4\) minutes, provided \(L>4\). The qualifying time has length \((L-4)_+=\max(L-4,0)\).
For the four successive lengths 3, 7, 2 and 8, the qualifying portions have lengths 0, 3, 0 and 4 minutes. These portions are disjoint within the 20-minute cycle. Uniform calendar-time sampling gives
\[\Pr(\text{remaining time}>4)=\frac{0+3+0+4}{3+7+2+8}=\frac7{20}.\]
Equivalently, the inspector enters the seven-minute interval with probability \(7/20\) and then has chance \(3/7\) of seeing more than four minutes left, or enters the eight-minute interval with probability \(8/20\) and then has chance \(4/8\). Their sum is \(3/20+4/20=7/20\). Counting the long intervals as two of four without weighting by their duration would answer a different sampling question. Strict versus non-strict boundaries do not alter a continuous-time probability.
Assumptions
The service schedule is deterministic and repeats without gaps or overlaps.
Arrival is uniform over the 20-minute cycle, independently of the schedule.
Remaining time is measured to the end of the interval containing the inspector's arrival.
Final Answer
\(\frac7{20}=35\%\).