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(A) Given,$P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$. We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\overline{A} \cap \o

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$0.5$

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(A) Given,$P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$. We need to find the probability that neither $A$ nor $B$ occurs,which is $P(\overline{A} \cap \overline{B})$. By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$. First,calculate $P(A \cup B)$ using the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.2 = 0.5$. Therefore,$P(\overline{A} \cap \overline{B}) = 1 - 0.5 = 0.5$.

If $A, B$ and $C$ are three events of a random experiment with $P(A)=0.4, P(B)=0.3$ and $P(A \cap B)=0.2$,then the probability that neither $A$ nor $B$ occurs is

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