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!
 A: $Pr (B|\neg A)$ 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 $Pr (A \cup B)$.
A: first,  the reason $Pr(B|¬A)  \ne Pr(B|A) $ is because $B|¬A$ is the event that $B$ happens if $A$ did not happen. And even though $A$ happening or not is the same, it doesn't mean that the next event after $A$ is the same. 
eg:
Think about it like this: say $A$ is tossing a head in a coin toss,  then I say if it's head we will roll one die, if it's tails we will roll two dies, and we let $B$ be the event that you get a 1 in your die rolling. Then $Pr(B|A) = 1/6$ and $Pr(B|¬A) =1/36$,while $Pr(A)=Pr(¬A) =0.5$. 
next your second question
We want $Pr(A \cup B) $ observe that
$$ not (A\cup B) = ¬A \cap ¬B $$
Note we can find $Pr(¬A \cap ¬B) $ by noticing that $$Pr(¬B |¬A) = 1-Pr(B|¬A) =2/5$$ which leads to $$ Pr(¬A \cap ¬B) = Pr(¬B|¬A)Pr(¬A) = 2/5 \cdot 1/2 = 1/5$$
So $$ Pr(A\cup B)  = 1-Pr(¬A \cap ¬B) =4/5$$
A: The  event $A\cup B$ is: $A\cup B=A\cup(B\setminus A)$.
Now you have $P(A\cup B)=P(A)+P(B\setminus A)-P(A\cap (B\setminus A))$.
Express $B\setminus A$ as an intersection and then replace in the last equality
$P(A)$, $P(B\setminus A)$ and $P(\emptyset)$.
A: Avoiding fractions and using directly the data:$$P\left(A\cup B\right)=P\left(A\cup B\mid A\right)P\left(A\right)+P\left(A\cup B\mid\neg A\right)P\left(\neg A\right)=$$$$1P\left(A\right)+P\left(B\mid\neg A\right)P\left(\neg A\right)=\frac{1}{2}+\frac{3}{5}(1-\frac{1}{2})=\frac{4}{5}$$
Note that under condition $\neg A$ the events $A\cup B$ and $B$ come to the same.
For $\omega\notin A$ "being element" of $A\cup B$ is the same as "being element" of $B$.
A: $P({B}|\neg{A})=\dfrac{3}{5}\implies$
$\dfrac{P({B}\cap\neg{A})}{P(\neg{A})}=\dfrac{3}{5}\implies$
$\dfrac{P({B}\cap\neg{A})}{(1-\frac{1}{2})}=\dfrac{3}{5}\implies$
$P({B}\cap\neg{A})=\dfrac{3}{10}$

$P({A}\cup{B})=$
$P(A)+\color{red}{P(B)}-P({A}\cap{B})=$
$P(A)+\color{red}{P({B}\cap({A}\cup\neg{A}))}-P({A}\cap{B})$
$P(A)+\color{red}{P({B}\cap{A})+P({B}\cap\neg{A})}-P({A}\cap{B})$
$P(A)+\color{red}{P({A}\cap{B})+P({B}\cap\neg{A})}-P({A}\cap{B})$
$P(A)+P({B}\cap\neg{A})=$
$\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{4}{5}$
