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(D) Let $A$ be the event of getting an even number on the first die. The possible outcomes for $A$ are $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1

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$\frac{20}{36}$

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(D) Let $A$ be the event of getting an even number on the first die. The possible outcomes for $A$ are $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$. Thus, $|A| = 18$.Let $B$ be the event of getting a sum of $8$. The possible outcomes for $B$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$. Thus, $|B| = 5$.The intersection $A \cap B$ contains outcomes where the first die is even and the sum is $8$, which are $(2,6), (4,4), (6,2)$. Thus, $|A \cap B| = 3$.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we get:$P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} = \frac{20}{36} = \frac{5}{9}$.

Two dice are tossed once. The probability of getting an even number on the first die or a total of $8$ is

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