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(B) Given that $P(A) = 0.5$ and $P(B) = 0.3$.Since $A$ and $B$ are mutually exclusive events,the probability of their union is given by $P(A \cup B) =

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$0.2$

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(B) Given that $P(A) = 0.5$ and $P(B) = 0.3$.Since $A$ and $B$ are mutually exclusive events,the probability of their union is given by $P(A \cup B) = P(A) + P(B) = 0.5 + 0.3 = 0.8$.We need to find the probability of happening neither $A$ nor $B$,which is $P(\overline{A} \cap \overline{B})$.By De Morgan's Law,$P(\overline{A} \cap \overline{B}) = P(\overline{A \cup B})$.Using the complement rule,$P(\overline{A \cup B}) = 1 - P(A \cup B)$.Therefore,the required probability is $1 - 0.8 = 0.2$.

The probability of happening an event $A$ is $0.5$ and that of $B$ is $0.3$. If $A$ and $B$ are mutually exclusive events,then the probability of happening neither $A$ nor $B$ is

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