JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 12)
Two dice are thrown independently. Let $$\mathrm{A}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is less than the number appeared on the $$2^{\text {nd }}$$ die, $$\mathrm{B}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is even and that on the second die is odd, and $$\mathrm{C}$$ be the event that the number appeared on the $$1^{\text {st }}$$ die is odd and that on the $$2^{\text {nd }}$$ is even. Then :
A and B are mutually exclusive
the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively
B and C are independent
the number of favourable cases of the event $$(\mathrm{A\cup B)\cap C}$$ is 6
Explanation
$\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$
$$ (4,5) \in A \text { and }(4,5) \in B $$
$\therefore A$ and $B$ are not exclusive events
$$ \begin{aligned} & n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\ = & 3+3-0 \\\\ = & 6 \end{aligned} $$
Option (D) is correct.
$$ \begin{aligned} & n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\ & \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\ & \therefore B \text { and } C \text { are not independent. } \end{aligned} $$
$$ (4,5) \in A \text { and }(4,5) \in B $$
$\therefore A$ and $B$ are not exclusive events
$$ \begin{aligned} & n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\ = & 3+3-0 \\\\ = & 6 \end{aligned} $$
Option (D) is correct.
$$ \begin{aligned} & n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\ & \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\ & \therefore B \text { and } C \text { are not independent. } \end{aligned} $$
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