Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 207 / 4 / a / ii

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- Solution ii

Solution

No. The high-dose expectation is

E_{B} = \frac{1}{2} + \frac{1}{2} + 3 \frac{4}{5} = 3.4.

Although both groups start with five patients, censoring removes two low-dose patients at month 3 and one high-dose patient at month 1. Their later risk sets therefore differ; only the sum

E_{A} + E_{B} = 5

must equal the total observed failures.

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