Question

An ordinary cube has 4 blank faces, one face marked 2 and another marked 3. Then the probability of obtaining 12 in 5 throws is:

• A. $\frac{5}{1296}$
• B. $\frac{5}{1944}$
• C. $\frac{5}{2592}$
• D. $None$

Answer 1

The correct option is A $\frac{5}{1296}$

$\begin{array}{cccc}Case:& 2& 3& Blanks\\ Case1:& 0& 4& 1\\ Case2:& 3& 2& 0\end{array}$
$P\left(Case1\right){:}^5{C}_4(\frac{1}{6}{)}^4.\frac{4}{6}=\frac{20}{6^5}P(Case2\left){:}^5{C}_3\right(\frac{1}{6}{)}^2.(\frac{1}{6}{)}^3=\frac{10}{6^5}.$
P(sum of 12): = P (case 1) + P (case 2)
$=\frac{20}{6^5}+\frac{10}{6^5}=\frac{30}{6^5}=\frac{5}{6^4}=\frac{5}{1296}$