Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 207 / 4 / b / Solution

Now treat the seventh observation as an actual event at $t_4$, not as a censoring at that time. The risk sets are $(7,6,5,4,2,1)$ and the event counts are $(1,1,1,2,1,1)$. The Kaplan–Meier estimator therefore givesConditional on reaching $t_4$, the two events at that time have total probability $2/4=1/2$. Surviving them has probability $1/2$, which is then divided equally between the remaining two event times. Thus