A six faced die is so biased that it is twice likely to show an even number as an odd number when thrown. If the die is thrown twice the probability that sum of the numbers is even is

Let $p$ denote the probability of showing an odd number, then the probability that die will show an even number is $2p$. We have

$p+2p=1\Rightarrow 3p=1\Rightarrow p=1/3$

Thus, probability of getting an even sum when dice is thrown twice is

$\begin{array}{c}P\left(\text{ odd, odd }\right)+P\left(\text{ even, even }\right)=P(\text{ odd }{)}^2+P(\text{ even }{)}^2\\ =p^2+(2p{)}^2=5p^2=5/9\end{array}$