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(C) The sample space of throwing a die is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.Event $A$ is the event that the num

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$1$

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(C) The sample space of throwing a die is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes is $n(S) = 6$.Event $A$ is the event that the number obtained is greater than $3$,so $A = \{4, 5, 6\}$. Thus,$n(A) = 3$ and $P(A) = 3/6$.Event $B$ is the event that the number obtained is less than $5$,so $B = \{1, 2, 3, 4\}$. Thus,$n(B) = 4$ and $P(B) = 4/6$.The intersection $A \cap B$ is the set of outcomes common to both $A$ and $B$,so $A \cap B = \{4\}$. Thus,$n(A \cap B) = 1$ and $P(A \cap B) = 1/6$.Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values,$P(A \cup B) = 3/6 + 4/6 - 1/6 = 6/6 = 1$.

$A$ die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $P(A \cup B)$ is

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