 Solution

An individual has neither died nor been censored by time $t$ exactly when $T>t$ and $C>t$. Under independent censoring, this has probability $F_T(t)G(t)$. Thus the probability of leaving observation by either route isThe two pieces agree at $t={\tau }_c-{\tau }_b$. The expression counts the first of death and censoring, so it does not double-count individuals who would eventually experience both.