Quadratic Equation Question 10
Question 10 - 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: Distributive Laws, Addition theorem of probability
$\begin{gathered} A={(1,2)(1,3)(1,4)(1,5)(1,6)(2,3)(2,4)(2,5)(2,6)(3,4)(3,5)(3,6)(4,5)(4,6)(5,6)} \\ B={(2,1)(2,3)(2,5)(4,1)(4,3)(4,5)(6,1)(6,3)(6,5)} \\ C={(1,2)(1,4)(1,6)(3,2)(3,4)(3,6)(5,2)(5,4)(5,6)} \\ (A \cup B) \cap C=(A \cap C) \cup(B \cap C) \\ =(A \cap C) \cup \phi \\ =A \cap C \\ \therefore n(A \cup B) \cap C)=n(A \cap C)=6 \end{gathered}$
