If A and B are independent events such that $Pr(A)=1/3$ and $Pr(B)>0$, what is the value of $Pr(A$ $\cup$ $B^c$$|B)=?$
From what I can understand , if we use the conditional probability formula , the numerator will be $Pr(A$ $\cup$ $B^c$ $\cap$ $B)$ which will be $0$ and therefore the answer is $0$ however im not sure. Can someone confirm or tell me where im going wrong?

$$P(A \cup B^c|B)\\=\frac{P(\left(A \cup B^c \right)\cap B)}{P(B)}\\=\frac{P(A\cap B)}{P(B)}\\=\frac{P(A)P(B)}{P(B)}\\=P(A)\\=\frac13$$
(The simplification at the second equality is obtained using either a Venn diagram or one of the distributive laws of set algebra.)