Answer
$0.903$
Work Step by Step
According to Bayes' theorem:
$P(C|D)=\dfrac{P(D|C)P(C)}{P(D|C)P(C)+P(D|C')+P(C')}~~~~~~~~(1)$
Here, we have
$P(D|C)= 0.7 \\ P(C')= 27.3 \% =0.273 \\ P(D|C')=0.2$
Now, we will now use formula (1) and the given data to obtain:
$P(C|D)=\dfrac{P(D|C)P(C)}{P(D|C)P(C)+P(D|C')+P(C')}=\dfrac{(0.7)(1-0.273)}{(0.7)(1-0.273)+(0.2) (0.273)}$
or, $ \approx 0.903$
Thus, we conclude the probability that the victim of a deadly side-impact accident was driving a car or a SUV is approximately $0.903$.