An ordinary cube has four blank faces,one fac | vedclass

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(C) The total number of outcomes in $5$ throws is $6^5 = 7776$.Let the faces be $0, 0, 0, 0, 2, 3$. We want a sum of $12$ in $5$ throws.Case $1$: One

Vedclass · Accepted Answer

$\frac{5}{2592}$

Vedclass · Answer

(C) The total number of outcomes in $5$ throws is $6^5 = 7776$.Let the faces be $0, 0, 0, 0, 2, 3$. We want a sum of $12$ in $5$ throws.Case $1$: One $0$ and four $3$'s. The sum is $0 + 3 + 3 + 3 + 3 = 12$. The number of ways is $\frac{5!}{1!4!} = 5$.Case $2$: Three $2$'s and two $3$'s. The sum is $2 + 2 + 2 + 3 + 3 = 12$. The number of ways is $\frac{5!}{3!2!} = 10$.Total favorable outcomes = $5 + 10 = 15$.Required probability = $\frac{15}{6^5} = \frac{15}{7776} = \frac{5}{2592}$.

An ordinary cube has four blank faces,one face marked $2$,and another marked $3$. The probability of obtaining a total of exactly $12$ in $5$ throws is:

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