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

At times 1 and 2, the risk sets immediately before failure contain respectively $(5,5)$ and $(4,4)$ low- and high-dose patients, with one failure each. The low-dose expected counts are $1/2$ and $1/2$. Immediately before time 5, one low-dose and four high-dose patients remain; three tied failures give low-dose expectation $3(1/5)=3/5$. Thus