In a bag there are three tickets numbered 1, 2, 3. A ticket is drawn at random and put back, and this is done four times the probability that the sum of the numbers is even is:

41/81

Let event O : odd no. of ticket ; E : even no. of ticket

The sum will be even for combinations : $\begin{matrix}O&O&O&O\\O&O&E&E\\E&E&E&E\end{matrix}→\frac{4!}{2!2!}$ ways

$P(O)=\frac{2}{3};P(E)=\frac{1}{3}$ ⇒ P(sum is even) = $(\frac{2}{3})^4+(\frac{2}{3})^2(\frac{1}{3})^2.\frac{4!}{2!2!}+(\frac{1}{3})^4=\frac{41}{81}$