Two dice are thrown independently. Let A be the event that the number appeared on the 1 st  die is less…

Given, $A$: No. on $1^{\text{st }}$die $<$ No. on $2^{\text{nd }}$ die

$A=\left\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\right\}$

$\therefore n\left(A\right)=15$

$B$: No. on $1^{\text{st }}$ die $=$even $\&$ No. of $2^{\text{nd }}$ die $=$ odd

$B=\left\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5), (6,1),(6,3),(6,5)\right\}$

$\therefore n\left(B\right)=9$

$C:$ No. on $1^{\text{st }}$ die $=$odd $\&$ No. on $2^{\text{nd }}$ die $=$even

$C=\left\{(1,2),(1,4),(1,6),(1,2),(1,4),(1,6),(5,2),(5,4),(5,6)\right\}$

$\therefore n\left(C\right)=9$

Now,
$n(A\cap B)=3, n(A\cap C)=6, n(B\cap C)=0$

$n(A\cap B\cap C)=0$

Since $(4,5)\in A$ and $(4,5)\in B$

$\therefore$ $A$ and $B$ are not exclusive events

Now,

$\mathrm{n}\left(\right(\mathrm{A}\cup \mathrm{B})\cap \mathrm{C})=n(\mathrm{A}\cap \mathrm{C})+n(\mathrm{B}\cap \mathrm{C})-n(A\cap B\cap C)$

$=6$

Since, $P\left(B\right)=\frac{9}{36}, P\left(C\right)=\frac{9}{36}, P(B\cap C)=0$

$\Rightarrow P\left(B\right).P\left(C\right)\ne P(B\cap C)$

$\therefore B$ and $C$ are not independent