Solution
= 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
$$
E_A=\frac12+\frac12+\frac35=1.6.
$$