Two dice are thrown and the sum of the number | vedclass

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(C) Let $A$ be the event that the sum of the numbers on the two dice is a multiple of $4$. The possible sums are $4, 8, 12$.The outcomes for event $A$

Vedclass · Accepted Answer

$\frac{7}{3}$

Vedclass · Answer

(C) Let $A$ be the event that the sum of the numbers on the two dice is a multiple of $4$. The possible sums are $4, 8, 12$.The outcomes for event $A$ are: $(1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4), (6, 6)$.Total number of outcomes in $A$ is $n(A) = 9$.Let $B$ be the event that the number $4$ appears at least once on the dice.We need to find the conditional probability $p = P(B|A) = \frac{n(A \cap B)}{n(A)}$.The intersection $A \cap B$ consists of outcomes where the sum is a multiple of $4$ $AND$ the number $4$ appears at least once.From the set $A$,the outcomes containing at least one $4$ are: $(4, 4)$.Thus,$n(A \cap B) = 1$.Therefore,$p = P(B|A) = \frac{1}{9}$.Finally,we calculate $3p + 2 = 3 	imes \frac{1}{9} + 2 = \frac{1}{3} + 2 = \frac{7}{3}$.

Two dice are thrown and the sum of the numbers appearing on the dice is observed to be a multiple of $4$. If $p$ is the conditional probability that number $4$ has appeared at least once,then $3p + 2 =$

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