# Probability question on union sets

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 .

• 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 $P(H\cup E^c)$. – 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 $$\{1,2,3,4,5,6\}$$