I have one probability question, which I got an answer different from the one given in the textbook. Here it is:
Suppose $A,B$ are two events in the same sample space $\Omega$, such that $\mathbb{P}(\overline{A})=0.3, \mathbb{P}(B)=0.4$ and $\mathbb{P}(A\cap\overline{B})=0.5$, calculate the conditional probability $\mathbb{P}(B|(A\cup\overline{B}))$. Here $\overline{A}$ denotes the opposite event of $A$, and similar for $\overline{B}$.
My way to solve the problem: By definition of conditional probability, we have $$\mathbb{P}(B|(A\cup\overline{B}))=\frac{\mathbb{P}(B\cap(A\cup\overline{B}))}{\mathbb{P}(A\cup\overline{B})}.$$
Note that $$B\cap(A\cup\overline{B})=(B\cap A)\cup(B\cap\overline{B})=B\cap A, $$ we have $$\mathbb{P}(B\cap(A\cup\overline{B}))=\mathbb{P}(B\cap A)=\mathbb{P}(A)-\mathbb{P}(A\cap\overline{B})=(1-0.3)-0.5=0.2.$$ On the other hand, $$\mathbb{P}(A\cup\overline{B})=\mathbb{P}(A)+\mathbb{P}(\overline{B})-\mathbb{P}(A\cap\overline{B})=(1-0.3)+(1-0.4)-0.5=0.8.$$ Hence $$\mathbb{P}(B|(A\cup\overline{B}))=\frac{0.2}{0.8}=0.25.$$
However, the answer given by the textbook is $\dfrac{1}{1960}$, without any justification and explanation. I wonder whether I made any mistakes or the textbook's answer is wrong. Any comments?