Probability Question 14
Question 14 - 01 February - Shift 2
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 $2^{\text{nd }} d i e, B$ 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
(1) the number of favourable cases of the event $(A \cup B) \cap C$ is 6
(2) A and B are mutually exchusive
(3) The number of favourable cases of the events
A, B and C are 15, 6 and 6 respectively
(4) B and C are independent
Show Answer
Answer: (1)
Solution:
Formula: Number of favourable outcomes, Distributive Laws, Mutually Exclusive events, Independent Events
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 $1^{\text{st }} d i e=$ odd and no. on $2^{\text{nd }} d i e=$ even
$n(A)=5+4+3+2+1=15$
$n(B)=9$
$n(C)=9$
$n((A \cup B) \cap C)=(A \cap C) \cup(B \cap C)$
$=(3+2+1)+0=6$.
