Two dice are thrown independently. Let A be the event that the...

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

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

$∴ n ( A ) = 15$

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

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

$∴ n ( B ) = 9$

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

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

$∴ n ( C ) = 9$

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 $(4,5) \in B$

$∴ A$ and $B$ are not exclusive events

Now,

$n ( ( A \cup B ) \cap C ) = n ( A \cap C ) + n ( B \cap C ) - n ( A \cap B \cap C )$

$= 6$

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

$⇒ P ( B ). P ( C ) \ne P ( B \cap C )$

$∴ B$ and $C$& are not independent