 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/12/f/solution

Use the lower limit

s = t + r_{0}

for the inner integral:

P (r > r_{0}) = \int_{0}^{\infty} \frac{e ^{- t / \overset{ˉ}{t}}}{t ˉ} [\int_{t + r_{0}}^{\infty} \frac{e ^{- s / \overset{s}{ˉ}}}{s ˉ} d s] d t = \frac{e ^{- r_{0} / \overset{s}{ˉ}}}{t ˉ} \int_{0}^{\infty} e^{- t (1/ \overset{s}{ˉ} + 1/ \overset{ˉ}{t})} d t .

Hence

P (r > r_{0}) = \frac{s ˉ}{s ˉ + t ˉ} e^{- r_{0} / \overset{s}{ˉ}}, r_{0} > 0.

The prefactor is the probability that interruption occurs; the remaining exponential is its conditional positive tail. This is the positive difference of independent exponential waiting times. The exponent is dimensionless because both

r_{0}

and

\overset{s}{ˉ}

are measured in minutes.

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