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

Vedclass · Accepted Answer

$\frac{11}{36}$

Vedclass · Answer

(B) When two dice are thrown simultaneously,the total number of outcomes is $n(S) = 6 	imes 6 = 36$. Let $A$ be the event of getting a multiple of $2$ on the first die,i.e.,$A \in \{2, 4, 6\}$. Let $B$ be the event of getting a multiple of $3$ on the second die,i.e.,$B \in \{3, 6\}$. We want to find the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other. Let $E$ be the set of favorable outcomes: $E = \{(2,3), (4,3), (6,3), (2,6), (4,6), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)\}$. Note that $(6,6)$ is included because it satisfies the condition of having a multiple of $2$ on one die and a multiple of $3$ on the other. Counting the elements in $E$,we get $n(E) = 11$. Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{11}{36}$.

Two dice are thrown simultaneously. The probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other die is

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