= Solution
Assign the seventh individual's event weights $(1/3,1/3,1/3)$, the conditional probabilities obtained from the original <Kaplan–Meier estimator>. The last three fractional counts are
$$
(r_4,d_4)=(4,4/3),\qquad(r_5,d_5)=(8/3,4/3),\qquad(r_6,d_6)=(4/3,4/3).
$$
Their conditional event fractions are $1/3$, $1/2$, and $1$, exactly as in the original censored-data calculation. Hence \b[the fractional allocation reproduces the original estimate]:
$$
\boxed{\widehat F(t_5)=\frac47\frac23\frac12=\frac4{21}=\widehat F_5^*.}
$$
This illustrates <self-consistency of Kaplan–Meier estimation>: distributing the unobserved event according to the current conditional fitted event distribution leaves the fitted distribution unchanged. The first imputation in part (b) was not such a self-consistent allocation.
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