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(B) When two dice are thrown,the total number of outcomes is $n(S) = 6 	imes 6 = 36$. Event $A$ is the event of getting a sum $\ge 8$. The outcomes a

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$\frac{6}{15}$

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(B) When two dice are thrown,the total number of outcomes is $n(S) = 6 	imes 6 = 36$. Event $A$ is the event of getting a sum $\ge 8$. The outcomes are: $A = \{(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)\}$. Thus,$n(A) = 15$. Event $B$ is the event of getting a number $\le 3$ on at least one die. Event $A \cap B$ is the event of getting a sum $\ge 8$ $AND$ at least one die showing a number $\le 3$. Looking at the elements of $A$,the outcomes where at least one die is $\le 3$ are: $A \cap B = \{(2,6), (3,5), (3,6), (5,3), (6,2), (6,3)\}$. Thus,$n(A \cap B) = 6$. The conditional probability is given by $P(B / A) = \frac{n(A \cap B)}{n(A)} = \frac{6}{15} = \frac{2}{5}$.

Two dice are thrown simultaneously. If $A$ is the event of getting the sum of the numbers on two dice as greater than or equal to $8$ and $B$ is the event of getting a number less than or equal to $3$ on at least one of the dice. Then,$P(B / A) = $

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