1. **State the problem:** Given probabilities $P(A) = 0.40$, $P(B) = 0.52$, and $P(\text{neither } A \text{ or } B) = 0.20$, find $P(A \cup B)$. 2. **Recall the formula:** The probability of the union of two events is given by $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ However, we are not directly given $P(A \cap B)$, so we need to use the information about neither event occurring. 3. **Use the complement rule:** The probability of neither $A$ nor $B$ occurring is the complement of $A \cup B$, so $$P(\text{neither } A \text{ or } B) = 1 - P(A \cup B)$$ Given $P(\text{neither } A \text{ or } B) = 0.20$, we have $$0.20 = 1 - P(A \cup B)$$ 4. **Solve for $P(A \cup B)$:** $$P(A \cup B) = 1 - 0.20 = 0.80$$ 5. **Interpretation:** The probability that either event $A$ or event $B$ (or both) occurs is $0.80$.