An arbitrary cube has four blank faces, one face marked 2 and another marked 3. Then the probability of obtaining a total of exactly 12 in 5 throws is

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$\frac{11}{1294}$

$\frac{5}{2592}$

$\frac{5}{1296}$

$\frac{5}{1944}$

answer is C.

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Detailed Solution

a die having $0, 0, 0, 0, 2, 3$

$1 2 3 4 5 3 3 2 2 2 3 3 3 3 0$

$\begin{array}{rcl} n (S) & = & 6^{5} \\ E & = & Sum is 12 \\ n (E) & = & 10 + 5 = 15 \\ P (E) & = & \frac{15}{6^{5}} = \frac{5}{6^{4} \times 2} = \frac{5}{2592} \end{array}$

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