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(C) We are given $P(A \cap \overline{B}) = 0.1$, $P(\overline{A} \cap B) = 0.2$, and $P(B) = 0.5$. Since $B = (A \cap B) \cup (\overline{A} \cap B)$,

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$0.4$

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(C) We are given $P(A \cap \overline{B}) = 0.1$, $P(\overline{A} \cap B) = 0.2$, and $P(B) = 0.5$. Since $B = (A \cap B) \cup (\overline{A} \cap B)$, and these are disjoint events, we have $P(B) = P(A \cap B) + P(\overline{A} \cap B)$. Substituting the values, $0.5 = P(A \cap B) + 0.2$, which gives $P(A \cap B) = 0.3$. Now, $P(A) = P(A \cap B) + P(A \cap \overline{B}) = 0.3 + 0.1 = 0.4$. We want to find $P(\overline{A} \cap \overline{B})$. By De Morgan's Law, $P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$. Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get $P(A \cup B) = 0.4 + 0.5 - 0.3 = 0.6$. Therefore, $P(\overline{A} \cap \overline{B}) = 1 - 0.6 = 0.4$.

Let $A$ and $B$ be two events in a random experiment. If $P(A \cap \overline{B}) = 0.1$, $P(\overline{A} \cap B) = 0.2$ and $P(B) = 0.5$, then $P(\overline{A} \cap \overline{B}) = $

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