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

0,0,0,1,2,3 are the numbers on 6 faces of a die where blank face denote a zero. To get sum 12 in 5 throws 
We can get (3,3,2,2,2) or (3,3,3,2,1) or (3,3,3,3,0) in any order, No. of favourable
$\begin{array}{l}\text{ Cases }{=}^5{\mathrm{C}}_2+\left({ }^5{\mathrm{C}}_3\times 2\right)+\left({ }^5{\mathrm{C}}_4\times 3\right)=45\\ [\therefore 0\text{ cab be obtained in }3\text{ ways }]\end{array}$
Required probability $=\frac{45}{\left(6^5\right)}$