Probablity identities and $A$ given $B$ I'm stuck on a question in my homework: 


  Assume that $\Pr(A) = 1/2$ and $\Pr(B\mid ¬A) = 3/5$. Determine $\Pr(A \cup B)$.


I'm assuming $\Pr(A)$ is equal to $\Pr(¬A)$ because $1-1/2 = 1/2$. I'm confused to why this doesn't mean $\Pr (B\mid ¬A) = \Pr(B\mid A)$ (according to my friend).  Assuming the previous isn't true, I got this far:


  $\Pr(¬A \cap B) =  3/5 \cdot 1/2$


Thanks for the help!
$Pr (B|\neg A)$ means "the probability of $B$ to happen given $A$ is false (or $A$ does not happen)." $B$ may be more likely to happen more if $A$ happens, $B$ may be less likely to happen if $A$ happens. Hence $Pr (B|\neg A)$ can have nothing to do with $Pr (B|A)$.

On the other hand, $Pr(A)$ does imply $Pr(\neg A)$, and $Pr (B|\neg A)$ does imply $Pr (\neg B|\neg A)$.

$Pr (A \cup B)$ means "the probability of $A$ or $B$ (or both) to happen". So you should find out the probability where both $A$ and $B$ do not happen (say $y$) by using $Pr(\neg A)$ and $Pr (\neg B|\neg A)$ found above, and then use $1-y$ to find out $Pr (A \cup B)$.
$P({B}|\neg{A})=\dfrac{3}{5}\implies$

$\dfrac{P({B}\cap\neg{A})}{P(\neg{A})}=\dfrac{3}{5}\implies$

$\dfrac{P({B}\cap\neg{A})}{(1-\frac{1}{2})}=\dfrac{3}{5}\implies$

$P({B}\cap\neg{A})=\dfrac{3}{10}$



$P({A}\cup{B})=$

$P(A)+\color{red}{P(B)}-P({A}\cap{B})=$

$P(A)+\color{red}{P({B}\cap({A}\cup\neg{A}))}-P({A}\cap{B})$

$P(A)+\color{red}{P({B}\cap{A})+P({B}\cap\neg{A})}-P({A}\cap{B})$

$P(A)+\color{red}{P({A}\cap{B})+P({B}\cap\neg{A})}-P({A}\cap{B})$

$P(A)+P({B}\cap\neg{A})=$

$\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{4}{5}$