In a single throw of 3 dice,find the probabil | vedclass

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(A) The total number of possible outcomes when throwing $3$ dice is $6 	imes 6 	imes 6 = 216$.To get a total of $18$,the only possible outcome is $(

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$\frac{1}{54}$

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(A) The total number of possible outcomes when throwing $3$ dice is $6 	imes 6 	imes 6 = 216$.To get a total of $18$,the only possible outcome is $(6, 6, 6)$,which is $1$ case.To get a total of $17$,the possible outcomes are $(5, 6, 6), (6, 5, 6), (6, 6, 5)$,which are $3$ cases.Therefore,the total number of favorable outcomes for a sum of $17$ or $18$ is $1 + 3 = 4$.The probability is given by $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{4}{216} = \frac{1}{54}$.

In a single throw of $3$ dice,find the probability of getting a total of $17$ or $18$.

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