Consider $P(H/E)>P(H)$. Is it true that $P(H\cup¬E/E)>P(H\cup ¬E)$.

I indicate with the negation sign the complement of a set .

  • $\begingroup$ You should know that "not E" is not going to happen when you are given $E$. The answer is you cannot say whether $P(H\mid E)$ is going to be larger than $P(H\cup E^c)$. $\endgroup$ – user10354138 Feb 23 at 15:23

Throw a fair die and consider $H=\{1,3,5\}$ being an odd value and $E=\{2,3,5\}$ a prime value.

Then $\mathbb P(H \mid E) = \frac23 > \frac12 = \mathbb P(H )$:

  • this is $\{3,5\}$ out of $\{2,3,5\}$, compared to $\{1,3,5\}$ out of $\{1,2,3,4,5,6\}$

but $\mathbb P(H \cup \lnot E \mid E) = \frac23 < \frac56 = \mathbb P(H \cup \lnot E)$:

  • this is $\{3,5\}$ out of $\{2,3,5\}$, compared to $\{1,3,4,5,6\}$ out of $\{1,2,3,4,5,6\}$

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