Eight minutes into a bus wait

Question

Buses on a certain route arrive at random, with the gap between buses exponentially distributed and averaging 12 minutes. You have already been waiting 8 minutes.

Given the 8 minutes already spent with no bus, what is the probability you must wait at least 6 more minutes?

Accepted Answer

The exponential is memoryless: the time already waited does not change the odds going forward. In symbols, P(T s + t \mid T s) = P(T t), which just says the wait effectively restarts from now. So the 8 minutes already elapsed drop out completely, and we only need the chance a fresh wait exceeds 6 minutes. With a mean gap of 12 minutes, the rate is \lambda = 	frac{1}{12} per minute. For an extra t = 6 minutes, P(	ext{at least 6 more}) = e^{-\frac{1}{12}\cdot 6} = e^{-0.5} \approx 0.607. It feels like you have "already put in your time," but the bus does not know that. Six more minutes is hal…

Eight minutes into a bus wait

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