If a die is rolled twice and the sum of the n | vedclass

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(B) Let $A$ be the event that the sum of the numbers appearing on the two dice is $6$. The possible outcomes for $A$ are: $(1, 5), (2, 4), (3, 3), (4,

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$\frac{2}{5}$

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(B) Let $A$ be the event that the sum of the numbers appearing on the two dice is $6$. The possible outcomes for $A$ are: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.Thus, the number of outcomes in $A$ is $n(A) = 5$.Let $B$ be the event that the number $1$ appears at least once.We are looking for the conditional probability $P(B|A)$, which is defined as $P(B|A) = \frac{n(A \cap B)}{n(A)}$.The intersection $A \cap B$ represents the outcomes where the sum is $6$ $AND$ the number $1$ appears at least once. These outcomes are: $(1, 5)$ and $(5, 1)$.Thus, $n(A \cap B) = 2$.Therefore, the required probability is $P(B|A) = \frac{2}{5}$.

If a die is rolled twice and the sum of the numbers appearing on them is observed to be $6$, then the probability that the number $1$ appears at least once on them is

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