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 :
Solution
$\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\\\ & n(A)=15 \\\\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\\\ & n(B)=9 \\\\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\\\ & n(C)=9\end{aligned}$
<br/><br/>$(4,5) \in A \text { and }(4,5) \in B$
<br/><br/>$\therefore A$ and $B$ are not exclusive events
<br/><br/>$$
\begin{aligned}
& n((A \cup B) \cap C)=n(A \cap C)+n(B \cap C)-n(A \cap B \cap C) \\\\
= & 3+3-0 \\\\
= & 6
\end{aligned}
$$
<br/><br/>Option (D) is correct.
<br/><br/>$$
\begin{aligned}
& n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\\\
& \Rightarrow n(B) \cdot n(C) \neq n(B \cap C) \\\\
& \therefore B \text { and } C \text { are not independent. }
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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