Conditional probability of a union of two independent events
Suppose $A$ & $B$ are independent events, what is the $P(B \mid A \cup B)$ ? In other words probability of $B$ given $A$ union $B$.
My idea was that $$\begin{align*}P(B\mid A \cup B) &= (P(B \cap (A \cup B))/ P(A \cup B) \\&= P(B)/P(A \cup B) \cap P(A \cup B) / P(A \cup B) \\&= P(B)/P(A \cup B) \cdot 1 \\&= P(B)/P(A \cup B)\end{align*}$$
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$\begingroup$Yes, your idea is correct. Applying Bayes rule, you have that $$P(B\mid A \cup B)=\frac{P(A \cup B \mid B)P(B)}{P(A\cup B)}=\frac{1\cdot P(B)}{P(A \cup B)}$$ where $P(A\cup B \mid B)=1$ since given that $B$ occured you are certain (i.e. prob$=1$) that $A$ or $B$ occured. Now the denominator can be further written as $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$$ since due to the independence of $A,B$ the identity $P(A\cap B)=P(A)P(B)$ holds.
In you answer, be careful that the symbols $\cup$ and $\cap$ are used as operators between sets (like $A$, $B$) and not between numbers or probabilities (like $P(A)$ or $P(B)$). Thus while it is correct to write $P(A \cup B)$ it is definitely wrong (and actually makes no sense) to write $P(A)\cup P(B)$. Instead use $+$ and $\cdot$ to make operations between probabilities, since they are numbers and not sets.
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