Two dice are tossed once. The probability of getting an even number at the first dice or a total of $8$ is

The correct option is D

$\frac{5}{9}$

Find the probability of the given condition:

The total number of outcomes$=36$

Let $A$be the event of getting an even number on the first dice and $B$ be the event of getting a total of $8$

Size of the event $A$$=\left\{\right(2,1),(2,2).......(2,6),(4,1),(4,2)......(4,6),(6,1),(6,2)......(6,6\left)\right\}$$=18$

Size of the event $B$ $=\left\{\right(2,6),(3,5),(4,4),(5,3),(6,2\left)\right\}=5$

Size of the event $A\cap B$$=\left\{\right(2,6),(4,4),(6,2\left)\right\}=3$

$\therefore P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{18}{36}+\frac{5}{36}-\frac{3}{36}$

$=\frac{20}{36}=\frac{5}{9}$

Hence, the correct option is (D).