A long call ending soon

Question

Call durations at a help line are exponentially distributed with a mean of 6 minutes. One call has already lasted 10 minutes and is still going.

Given that it has already run 10 minutes, what is the probability it ends within the next 3 minutes?

Accepted Answer

The exponential is memoryless, so the 10 minutes already spent on the call do not change what happens next. The remaining time behaves like a brand-new exponential wait: P(T s + t \mid T s) = P(T t). We want the chance the call ends within the next 3 minutes, which is the complement of lasting at least 3 more minutes. First the survival part for a fresh wait, with mean 6 minutes so \lambda = 	frac{1}{6} per minute and t = 3: P(	ext{at least 3 more}) = e^{-\frac{1}{6}\cdot 3} = e^{-0.5} \approx 0.607. Then the chance it ends within 3 minutes is what remains: P(	ext{ends within 3 more}) = 1 …

A long call ending soon

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