A die is cast such that it is twice more likely to show an even number than to show an odd number. The dice is thrown 3 times. Then the probability that the sum on them is 14 is

P(E)=2P(O)=k where P(E)=P( 2 or 4 or 6 ) and P(O)=P(1 or 3 or 5)

But P(E)+P(O)=1 $\Rightarrow$ k = 2/9 

$\therefore$P(E) = 2 / 3 &P(0) = 1 $\Rightarrow$P(2) = P(4) = P(6) = 2 / 9 and P(1) = P(3) = P(5) = 1/ 9 

To get sum 14 in 3 throws we can get (6,6,2), (6,5,3), (6,4,4), (5,5,4) in any order.
 

$\therefore$Req.Prob $=2\left[3\cdot {\left(\frac{2}{9}\right)}^3\right]+[6+3]\left[\left(\frac{2}{9}\right){\left(\frac{1}{9}\right)}^2\right]=\frac{22}{243}$