 Solution

ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-2/5/a/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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