Answer
$P(only~property~damage)=\frac{641}{922}\approx0.6952$
Work Step by Step
For the camera system:
The sample space: 922 crashes. So, $N(S)=922$
$N(reported~property~damage~only)=333$
$N(unreported~property~damage~only)=308$
The events "reported property damage only" and "unreported property damage only" are mutually exclusive (disjoint events).
Using the Addition Rule for Disjoint Events (page 270) and the Classical Method (page 259):
$P(only~property~damage)=P(reported~property~damage~only~or~unreported~property~damage~only)=\frac{N(reported~property~damage~only)}{N(S)}+\frac{N(unreported~property~damage~only)}{N(S)}=\frac{333}{922}+\frac{308}{922}=\frac{641}{922}\approx0.6952$
