An ordinary cube has 3 blank faces other 3 faces are marked with numbers 1, 2, 3. Then the probability of getting 12 in 5 throws is

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$\frac{50}{6^{5}}$

$\frac{40}{6^{5}}$

$\frac{45}{6^{5}}$

$\frac{25}{6^{5}}$

answer is A.

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Detailed Solution

$n (s) = 6^{5}$
E=sum 12 on 5 throws $= (3, 3, 3, 3 -) \to^{3} C_{1} \times \frac{5!}{4!} = 15$
$(3, 3, 3, 1, 2) \to \frac{5!}{3!} = 20$
$(3, 3, 2, 2, 2) \to \frac{5!}{2! 3!} = 10 \Rightarrow n (E) = 45$

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