$\{A,B,C\}$ are independent events and if $\Pr(A) = \frac{1}{2}$ , $\Pr(B) = \frac{1}{4}$ and $\Pr(C) = \frac{1}{5}$, then what is the probability that either $A$ or $B$ occur, not both.
I got $\frac {5}{8}$ as my answer but my professor got $\frac{1}{2}$ which I don't understand considering I used the correct formula for finding this if $A$ and $B$ are independent events
$$\Pr(A \text{ disjoint } B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$ and since the events are independent, then $$\Pr(A \cap B) = \Pr(A) \times \Pr(B).$$
You have computed the probability that $P(A \cup B)=P(A)+P(B)-P(A \cap B)$.
What you are interested would be $P(A \cup B)-P(A \cap B)=\frac58 -\frac18=\frac12$
The event of interest is $A \setminus (A \cap B)$ and $B \setminus (A \cap B)$ which are disjoint, hence the probability would be $$P(A \setminus (A \cap B))+P(B \setminus (A \cap B))=P(A)-P(A \cap B) + P(B)-P(A \cap B) = P( A \cup B)-P(A \cap B)$$