Two dice are thrown independently. Let $A$ be the event that the number appeared on 1st die is less than the number appeared on the 2nd die, $B$ be the event that the number appeared on the 1st die is even and that on the second die is odd, and $C$ be the event that the number appeared on the 1st die is odd and that on the 2nd is even. Then :

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$B$ and $C$ are independent

$A a n d B$ are mutually exclusive

The number of favourable cases of the events $A, B a n d C$ are 15, 6 and 6 respectively

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

answer is C.

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Detailed Solution

$\begin{array}{l} A = {(1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6)} \\ B = {(2, 1) (2, 3) (2, 5) (4, 1) (4, 3) (4, 5) (6, 1) (6, 3) (6, 5)} \\ C = {(1, 2) (1, 4) (1, 6) (3, 2) (3, 4) (3, 6) (5, 2) (5, 4) (5, 6)} \\ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) \\ = (A \cap C) \cup ϕ \\ = A \cap C \\ ∴ n ((A \cup B) \cap C) = n (A \cap C) = 6 \end{array}$

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