Simple conditional probability problem Question states: Consider two events A and B in a sample space S.
• Assume that Pr(A) = 1/2 and Pr(B|A') = 3/5. Determine Pr(A∪B).

Not quite sure how to answers this question. Here's how I started:

So Pr(A) = 1/2 so Pr(A') = 1/2. 

Then Pr(B|A) = 3/5.

Pr(A∩B) = Pr(A) * Pr(B|A) = 1/2 * 3/5 = 0.3

Pr(A∩B) = 0.3

But how do I get (A∪B)?

Pr(A∪B) = Pr(A) + Pr(B) - Pr(A∩B)

but I don't know the Pr(B)? What am I missing here? Thanks for any insight.
$P({B}|\neg{A})=\dfrac{3}{5}\implies$

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

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

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



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

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

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

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

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

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

$\dfrac{1}{2}+\dfrac{3}{10}=\dfrac{4}{5}$
$P({B}|\neg{A})=\dfrac{3}{5}\implies$

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

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

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



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

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

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

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

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

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

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