A dice marked with digits {1, 2, 2, 3, 3, 3} | vedclass

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(B) The probabilities of getting each digit on a single throw are:$P(1) = \frac{1}{6}$$P(2) = \frac{2}{6}$$P(3) = \frac{3}{6}$To get a sum of $6$ in t

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$\frac{44}{216}$

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(B) The probabilities of getting each digit on a single throw are:$P(1) = \frac{1}{6}$$P(2) = \frac{2}{6}$$P(3) = \frac{3}{6}$To get a sum of $6$ in three throws,the possible combinations of digits are:$1)$ $(1, 2, 3)$ in any order: The number of permutations is $3! = 6$. The probability for one such sequence is $\frac{1}{6} 	imes \frac{2}{6} 	imes \frac{3}{6} = \frac{6}{216}$. Total probability for this set is $6 	imes \frac{6}{216} = \frac{36}{216}$.$2)$ $(2, 2, 2)$: The number of permutations is $1$. The probability is $\frac{2}{6} 	imes \frac{2}{6} 	imes \frac{2}{6} = \frac{8}{216}$.Total probability $= \frac{36}{216} + \frac{8}{216} = \frac{44}{216}$.

$A$ dice marked with digits $\{1, 2, 2, 3, 3, 3\}$ is thrown three times. The probability that the sum of the numbers on the faces is $6$ is equal to:

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