1. **Understanding the problem:** We have mutually exclusive events $A$ and $B$ in the sample space $2^\Omega$. We want to find the probability of $\overline{A} \cap \overline{B}$, which means the event that neither $A$ nor $B$ occurs. 2. **Recall definitions:** - Since $A$ and $B$ are mutually exclusive, $A \cap B = \emptyset$. - The complement $\overline{A}$ is the event that $A$ does not occur. - Similarly, $\overline{B}$ is the event that $B$ does not occur. - The event $\overline{A} \cap \overline{B}$ means both $A$ and $B$ do not occur. 3. **Use set identity:** We know De Morgan's law: $$\overline{A} \cap \overline{B} = \overline{A \cup B}$$ 4. **Calculate the probability:** Since $A$ and $B$ are mutually exclusive, $$P(A \cup B) = P(A) + P(B)$$ Therefore, $$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B)]$$ 5. **Check given options:** - $P(A)$ is just the probability of $A$. - $P(B)$ is just the probability of $B$. - $P(\overline{A}) = 1 - P(A)$. - $P(\overline{B}) = 1 - P(B)$. None of these match $1 - [P(A) + P(B)]$ directly unless $P(A) + P(B) = 1$, which is not necessarily stated. **Final conclusion:** $P(\overline{A} \cap \overline{B}) = 1 - P(A) - P(B)$, which is not one of the options 1, 2, 3, or 4 unless further assumptions are made. So the answer is that $P(\overline{A} \cap \overline{B})$ is not equal to any of the four options provided unless $P(A)$ or $P(B)$ are zero or together sum to 1.