Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2025 / Paper 2 / 5 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The Waiting times of a Poisson process have an exponential distribution with the arrival rate. Measuring time in days, the survival function is

P (W > w) = e^{- α w}

. Therefore

P (W > 1) = e^{- 3} \approx 0.0497871,

or about $4.98%$ .

ii

Solution

A day's admission count is a Poisson random variable with mean three. “Less than three” includes zero, one and two, so

P (A < 3) = k = 0 \sum 2 \frac{3 ^{k} e ^{- 3}}{k !} = e^{- 3} (1 + 3 + \frac{9}{2}) = \frac{17}{2} e^{- 3} \approx 0.423190 .

Thus the probability is about $42.32%$ .

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