Total number of outcomes when two dice are thrown $=6 \times 6$
$\therefore n(S)=36$
Let $E=$ outcomes in which product of two number is even
$=\{(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3)(6,4),(6,5),(6,6)\}$
$\therefore n(E)=27$
$\therefore $ Required probability $=P(E)$
$=\frac{n(E)}{n(S)}$
$=\frac{27}{36}=\frac{3}{4}$