In a single throw of 2 dice,what is the proba | vedclass

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

Vedclass · Accepted Answer

$\frac{11}{36}$

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(C) When $2$ dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.Let $A$ be the event of getting a multiple of $2$ on the first die: $A = \{2, 4, 6\}$.Let $B$ be the event of getting a multiple of $3$ on the second die: $B = \{3, 6\}$.We need to find the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other.Let $E_1$ be the set of outcomes where the first die is a multiple of $2$ and the second is a multiple of $3$: $E_1 = \{(2,3), (2,6), (4,3), (4,6), (6,3), (6,6)\}$.Let $E_2$ be the set of outcomes where the first die is a multiple of $3$ and the second is a multiple of $2$: $E_2 = \{(3,2), (3,4), (3,6), (6,2), (6,4), (6,6)\}$.The union $E_1 \cup E_2$ represents the favourable outcomes. Note that $(6,6)$ is common to both sets.$E_1 \cup E_2 = \{(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (6,2), (6,4)\}$.Counting the unique elements,the number of favourable outcomes is $11$.Therefore,the required probability is $\frac{11}{36}$.

In a single throw of $2$ dice,what is the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other?

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