Two dice are thrown independently.

Question

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 :

Accepted Answer

$\begin{aligned} & A=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\} \\ & n(A)=15 \\ & B=\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\} \\ & n(B)=9 \\ & C=\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\} \\ & n(C)=9\end{aligned}$

$$
(4,5) \in A \text { and }(4,5) \in B
$$

$\therefore A$ 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.

$$
\begin{aligned}
& n(B)=\frac{9}{36}, n(C)=\frac{9}{36}, n(B \cap C)=0 \\
& \Rightarrow  n(B) \cdot n(C) \neq n(B \cap C) \\
& \therefore  B \text { and } C \text { are not independent. }
\end{aligned}
$$

Solutions · 1