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(B) The total number of outcomes when a pair of dice is thrown is $6 	imes 6 = 36$. To get a score of $9$,the possible outcomes are $(3, 6), (4, 5),

Vedclass · Accepted Answer

$\frac{8}{243}$

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(B) The total number of outcomes when a pair of dice is thrown is $6 	imes 6 = 36$. To get a score of $9$,the possible outcomes are $(3, 6), (4, 5), (5, 4), (6, 3)$. Thus,the number of favorable outcomes is $4$. The probability of getting a score of $9$ in a single throw is $p = \frac{4}{36} = \frac{1}{9}$. The probability of not getting a score of $9$ is $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$. Using the binomial distribution formula $P(X = k) = nC_k 	imes p^k 	imes q^{n-k}$,where $n = 3$ and $k = 2$: $P(X = 2) = 3C_2 	imes (\frac{1}{9})^2 	imes (\frac{8}{9})^{3-2}$ $= 3 	imes \frac{1}{81} 	imes \frac{8}{9}$ $= \frac{24}{729} = \frac{8}{243}$.

$A$ pair of fair dice is thrown independently three times. The probability of getting a score of exactly $9$ twice is

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