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

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(C) The total number of outcomes in $5$ throws of a $6$-faced die is $6^5 = 7776$.Let $X$ be the outcome of a single throw. The possible values are $0

Vedclass · Accepted Answer

$\frac{5}{2592}$

Vedclass · Answer

(C) The total number of outcomes in $5$ throws of a $6$-faced die is $6^5 = 7776$.Let $X$ be the outcome of a single throw. The possible values are $0$ (blank),$2$,and $3$.We want the sum of $5$ throws to be $12$.Let $n_0, n_2, n_3$ be the number of times $0, 2, 3$ appear respectively,where $n_0 + n_2 + n_3 = 5$.The sum is $0(n_0) + 2(n_2) + 3(n_3) = 12$.Case $1$: If $n_0 = 1$,then $2n_2 + 3n_3 = 12$. If $n_3 = 4$,then $2n_2 = 0 \implies n_2 = 0$. This gives $(n_0, n_2, n_3) = (1, 0, 4)$.The number of ways is $\frac{5!}{1!0!4!} = 5$.Case $2$: If $n_0 = 0$,then $2n_2 + 3n_3 = 12$. If $n_3 = 2$,then $2n_2 = 6 \implies n_2 = 3$. This gives $(n_0, n_2, n_3) = (0, 3, 2)$.The number of ways is $\frac{5!}{0!3!2!} = 10$.Total favorable outcomes = $5 + 10 = 15$.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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