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(A) Let the outcomes of the two dice be $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 	imes 6 = 36$.Event $A$: $

Vedclass · Accepted Answer

the number of favourable cases of the event $(A \cup B) \cap C$ is $6$

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(A) Let the outcomes of the two dice be $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $6 	imes 6 = 36$.Event $A$: $x Event $B$: $x \in \{2, 4, 6\}$ and $y \in \{1, 3, 5\}$. The number of outcomes is $3 	imes 3 = 9$.Event $C$: $x \in \{1, 3, 5\}$ and $y \in \{2, 4, 6\}$. The number of outcomes is $3 	imes 3 = 9$.Now,$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$.$B \cap C$: $x$ is even and odd,which is impossible,so $B \cap C = \emptyset$.$A \cap C$: $x If $x=1$,$y \in \{2, 4, 6\}$ ($3$ cases).If $x=3$,$y \in \{4, 6\}$ ($2$ cases).If $x=5$,$y \in \{6\}$ ($1$ case).Total cases for $A \cap C = 3 + 2 + 1 = 6$.Thus,the number of favourable cases for $(A \cup B) \cap C$ is $6 + 0 = 6$.

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