- Home
- Standard 11
- Mathematics
Two dice are thrown independently. Let $A$ 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, $B$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $C$ 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
the number of favourable cases of the event $(A \cup B) \cap C$ is $6$
$A$ and $B$ are mutually exchusive
The number of favourable cases of the events $A , B$ and $C$ are $15,6$ and $6$ respectively
$B$ and $C$ are independent
Solution
$A$ : no. on $1^{\text {st }}$ die < no. on $2^{\text {nd }}$ die
$A$ : no. on $1^{\text {st }}$ die $=$ even and no. of $2^{\text {nd }}$ die $=$ odd
$C :$ no. on $1^{\text {ti }}$ die $=$ odd and no. on $2^{\text {nd }} d i e=$ even
$n ( A )=5+4+3+2+1=15$
$n ( B )=9$
$n ( C )=9$
$n (( A \cup B ) \cap C )=( A \cap C ) \cup( B \cap C )$
$=(3+2+1)+0=6$.
