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(B) Let $A$ be the event of getting a sum of $9$ and $B$ be the event of getting a sum of $11$.The sample space for throwing $2$ dice has $6 	imes 6

Vedclass · Accepted Answer

$\frac{5}{6}$

Vedclass · Answer

(B) Let $A$ be the event of getting a sum of $9$ and $B$ be the event of getting a sum of $11$.The sample space for throwing $2$ dice has $6 	imes 6 = 36$ outcomes.The outcomes for event $A$ (sum $= 9$) are: $(3, 6), (4, 5), (5, 4), (6, 3)$. So,$n(A) = 4$.The outcomes for event $B$ (sum $= 11$) are: $(5, 6), (6, 5)$. So,$n(B) = 2$.Since $A$ and $B$ are mutually exclusive events,$P(A \cup B) = P(A) + P(B) = \frac{4}{36} + \frac{2}{36} = \frac{6}{36} = \frac{1}{6}$.The probability that the sum is neither $9$ nor $11$ is $1 - P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.

$2$ dice are thrown together. What is the probability that the sum of the numbers on $2$ faces is neither $9$ nor $11$?

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