P(numar par) =1-P(impar)
P impar trebuie ca pe ambele zarurisa fiecate un numar impar, adica 1;3sau5
evenimente independente total 3*3 variante cazuri favorabile
cazuri posibile 6*6=36
P impar =3 *3/36=9/36=1/4
p(numar par)=1-1/4
P(numar par) =3/4=75%
as simple as that!!!
