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(B) The sample space $S$ for rolling two dice has $6 	imes 6 = 36$ outcomes. Event $A$ is the event that the same number shows on each die: $A = \{(1

Vedclass · Accepted Answer

$\frac{1}{5}, \frac{1}{2}$

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(B) The sample space $S$ for rolling two dice has $6 	imes 6 = 36$ outcomes. Event $A$ is the event that the same number shows on each die: $A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$. Thus,$n(A) = 6$ and $P(A) = \frac{6}{36} = \frac{1}{6}$. Event $B$ is the event that the sum of the numbers is greater than $7$: $B = \{(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)\}$. Thus,$n(B) = 15$ and $P(B) = \frac{15}{36} = \frac{5}{12}$. The intersection $A \cap B$ is the set of outcomes where the numbers are the same and the sum is greater than $7$: $A \cap B = \{(4,4), (5,5), (6,6)\}$. Thus,$n(A \cap B) = 3$ and $P(A \cap B) = \frac{3}{36} = \frac{1}{12}$. Now,calculate the conditional probabilities: $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{5/12} = \frac{1}{5}$. $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/12}{1/6} = \frac{1}{2}$.

Two dice are rolled. If $A$ denotes the event that the same number shows on each die and $B$ denotes the event that the sum of the numbers on both dice is greater than $7$,then $P(A \mid B)$ and $P(B \mid A)$ respectively are

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