A die is thrown. Let A be the event that the | vedclass

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

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

$A$ 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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