Given that a throw of three unbiased dice sho | vedclass

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Question

(A) The total number of outcomes when three unbiased dice are thrown such that they show different faces is given by the number of ways to choose $3$

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(A) The total number of outcomes when three unbiased dice are thrown such that they show different faces is given by the number of ways to choose $3$ distinct numbers from $6$ and arrange them,which is $P(6, 3) = 6 	imes 5 	imes 4 = 120$. Alternatively,if we consider the set of faces as an unordered combination,there are $^6C_3 = 20$ sets of three distinct faces. To find the number of sets of three distinct faces that sum to $8$,we list the combinations: $(1, 2, 5)$ and $(1, 3, 4)$. There are $2$ such sets. Since each set of $3$ distinct faces can be arranged in $3! = 6$ ways,the total number of favorable outcomes is $2 	imes 6 = 12$. The total number of outcomes with distinct faces is $120$. Therefore,the required probability is $\frac{12}{120} = \frac{1}{10}$. Hence,option $A$ is correct.

Given that a throw of three unbiased dice shows different faces,what is the probability that their total is eight?

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