Big Nate: In The Zone And The Case Of The Missing Clues By Lincoln. Now, the legal action has come to an end and the dummies have been handed. Hootie in the Hall of Fame Book 2: Hootie’s Blues .
Why does $p(X|H) = 1$ imply $p(H|X) = 0$, if $H$ and $X$ are random variables?
In our lecture, we stated that $p(H|X) = 0$ implies $p(X|H) = 1$, but I did not understand why this is so.
Let $\underline{H}$ be the event «H is true,» $\underline{X}$ be the event «X is true,» and $\bar{H}$ be the event «H is false,» and $\bar{X}$ be the event «X is false.» Then $P(H|X) = P(\underline{H}|\underline{X}) = P(\bar{H}|\bar{X})$, by the definition of conditional probability. Now write $P(\underline{H}|\underline{X})$ as $P(\underline{H},\underline{X})/P(\underline{X})$ and similarly $P(\bar{H}|\bar{X})$ as $P(\bar{H},\bar{X})/P(\bar{X})$. Then
 P(\underline{H}|\underline{X}) = P(\bar{H}|\bar{X}) = \frac{P(\bar{H},\bar{X}) + P(\underline{H},\underline{X})}{P(\bar{X})+P(\underline{X})} = \frac{P(\bar{H},\bar{X}) + P(\underline{H},\underline{X})}{P(\bar{X})+P(\underline{