Solution
= Solution
Of the ten outcomes in $F$, all except $(1,1,1)$ have exactly one repeated pair. Hence $P(T\mid F)=9/10$. Applying <Bayes' theorem> in the requested direction,
$$
\boxed{P(F\mid T)=\frac{P(T\mid F)P(F)}{P(T)}=\frac{(9/10)(10/216)}{90/216}=\frac1{10}.}
$$
Equivalently there are nine favorable outcomes among the ninety outcomes in $T$. This direct count confirms the <conditional probability> calculation.