1. **State the problem:** We are given probabilities $P(A) = 0.40$, $P(B) = 0.52$, and $P(\text{neither } A \text{ nor } B) = 0.20$. We need to find the probability of the intersection $P(A \cap B)$. 2. **Recall the formula:** The probability of the union of two events $A$ and $B$ is given by: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$ 3. **Important rule:** The probability of neither $A$ nor $B$ happening is the complement of the union: $$ P(\text{neither } A \text{ nor } B) = 1 - P(A \cup B) $$ 4. **Calculate $P(A \cup B)$:** $$ P(A \cup B) = 1 - P(\text{neither } A \text{ nor } B) = 1 - 0.20 = 0.80 $$ 5. **Use the union formula to find $P(A \cap B)$:** $$ 0.80 = 0.40 + 0.52 - P(A \cap B) $$ 6. **Solve for $P(A \cap B)$:** $$ P(A \cap B) = 0.40 + 0.52 - 0.80 = 0.92 - 0.80 = 0.12 $$ 7. **Final answer:** $$ P(A \cap B) = 0.12 $$ This means the probability that both events $A$ and $B$ occur simultaneously is 0.12.