Answer
$P(E) + P(F)$
Work Step by Step
If $E$ and $F$ are disjoint events, they have no outcomes in common, that is: $N(E~or~F)=N(E)+N(F)$
Using the Classical Method (see page 259):
$P(E~or~F)=\frac{N(E~or~F)}{N(S)}=\frac{N(E)+N(F)}{N(S)}=\frac{N(E)}{N(S)}+\frac{N(F)}{N(S)}=P(E)+P(F)$
