= 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 gives
$$
\boxed{\widehat F_5^0=\frac47\left(1-\frac24\right)\left(1-\frac12\right)=\frac17.}
$$
Conditional 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
$$
\boxed{(\widehat p_4^0,\widehat p_5^0,\widehat p_6^0)=(1/2,1/4,1/4).}
$$
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