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(B) The total number of outcomes when throwing two dice is $6 	imes 6 = 36$. The outcomes resulting in a sum of $5$ are $(1, 4), (2, 3), (3, 2), (4,

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$123$

Vedclass · Answer

(B) The total number of outcomes when throwing two dice is $6 	imes 6 = 36$. The outcomes resulting in a sum of $5$ are $(1, 4), (2, 3), (3, 2), (4, 1)$,which are $4$ outcomes. The probability of success $p = \frac{4}{36} = \frac{1}{9}$. The probability of failure $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$. We use the binomial distribution formula $P(X = r) = {}^nC_r p^r q^{n-r}$ with $n = 5$. $P(	ext{at least } 4 	ext{ successes}) = P(X = 4) + P(X = 5)$. $P(X = 4) = {}^5C_4 	imes (\frac{1}{9})^4 	imes (\frac{8}{9})^1 = 5 	imes \frac{1}{9^4} 	imes \frac{8}{9} = \frac{40}{9^5} = \frac{40}{3^{10}}$. $P(X = 5) = {}^5C_5 	imes (\frac{1}{9})^5 	imes (\frac{8}{9})^0 = 1 	imes \frac{1}{9^5} = \frac{1}{3^{10}}$. Total probability $= \frac{40}{3^{10}} + \frac{1}{3^{10}} = \frac{41}{3^{10}} = \frac{41 	imes 3}{3^{11}} = \frac{123}{3^{11}}$. Comparing this with $\frac{k}{3^{11}}$,we get $k = 123$.

Two dice are thrown $5$ times,and each time the sum of the numbers obtained being $5$ is considered a success. If the probability of having at least $4$ successes is $\frac{k}{3^{11}}$,then $k$ is equal to

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