If two dice are rolled, then the probability | vedclass

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(C) Let $A$ be the event that the sum is a multiple of $3$, so $A = \{3, 6, 9, 12\}$.Let $B$ be the event that the sum is an odd number, so $B = \{3,

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$\frac{1}{3}$

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(C) Let $A$ be the event that the sum is a multiple of $3$, so $A = \{3, 6, 9, 12\}$.Let $B$ be the event that the sum is an odd number, so $B = \{3, 5, 7, 9, 11\}$.We need to find the conditional probability $P(A|B) = \frac{n(A \cap B)}{n(B)}$.For the sum to be both a multiple of $3$ and odd, the sum must be $3$ or $9$.The pairs $(x, y)$ such that $x+y = 3$ are $(1, 2)$ and $(2, 1)$.The pairs $(x, y)$ such that $x+y = 9$ are $(3, 6), (6, 3), (4, 5), (5, 4)$.Thus, $A \cap B = \{(1, 2), (2, 1), (3, 6), (6, 3), (4, 5), (5, 4)\}$ and $n(A \cap B) = 6$.The total number of outcomes where the sum is odd $(B)$ is $18$ (since exactly half of the $36$ outcomes result in an odd sum).Therefore, $P(A|B) = \frac{6}{18} = \frac{1}{3}$.

If two dice are rolled, then the probability of getting a multiple of $3$ as the sum of the numbers appeared on the top faces of the dice, given that their sum is an odd number, is

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