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(B) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$.Let $A$ be the event of getting a multiple of $2$ on the first die: $A

Vedclass · Accepted Answer

$\frac{11}{36}$

Vedclass · Answer

(B) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$.Let $A$ be the event of getting a multiple of $2$ on the first die: $A = \{2, 4, 6\}$,so $n(A) = 3$.Let $B$ be the event of getting a multiple of $3$ on the second die: $B = \{3, 6\}$,so $n(B) = 2$.The number of outcomes where the first die is a multiple of $2$ and the second is a multiple of $3$ is $3 	imes 2 = 6$. These are $(2,3), (2,6), (4,3), (4,6), (6,3), (6,6)$.Similarly,the number of outcomes where the first die is a multiple of $3$ and the second is a multiple of $2$ is $2 	imes 3 = 6$. These are $(3,2), (3,4), (3,6), (6,2), (6,4), (6,6)$.The intersection of these two sets is the outcome $(6,6)$,which is counted in both cases.Using the Principle of Inclusion-Exclusion,the total number of favourable outcomes is $6 + 6 - 1 = 11$.Therefore,the required probability is $\frac{11}{36}$.

Two dice are thrown simultaneously. What is the probability of obtaining a multiple of $2$ on one of them and a multiple of $3$ on the other?

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