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

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(B) The probabilities for the outcomes on a single throw are: $P(1) = \frac{1}{6}$,$P(2) = \frac{2}{6}$,and $P(3) = \frac{3}{6}$.To get a sum of $6$ i

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$\frac{44}{216}$

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(B) The probabilities for the outcomes on a single throw are: $P(1) = \frac{1}{6}$,$P(2) = \frac{2}{6}$,and $P(3) = \frac{3}{6}$.To get a sum of $6$ in three throws,the possible combinations of numbers are:$1)$ $\{1, 2, 3\}$ in any order.$2)$ $\{2, 2, 2\}$.For the combination $\{1, 2, 3\}$,the number of permutations is $3! = 6$. The probability is $6 	imes (P(1) 	imes P(2) 	imes P(3)) = 6 	imes (\frac{1}{6} 	imes \frac{2}{6} 	imes \frac{3}{6}) = \frac{36}{216}$.For the combination $\{2, 2, 2\}$,the number of permutations is $1$. The probability is $1 	imes (P(2) 	imes P(2) 	imes P(2)) = (\frac{2}{6} 	imes \frac{2}{6} 	imes \frac{2}{6}) = \frac{8}{216}$.The total probability is $\frac{36}{216} + \frac{8}{216} = \frac{44}{216}$.

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

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