Answer
$0.735$
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
$P(D|C)= 1 \\ P(C)= 45.4 \% =0.454 \\ P(D|C')=0.3$
Now, we will now use formula (1) and the given data to obtain:
$P(C|D)=\dfrac{P(D|C)P(C)}{P(D|C)P(C)+P(D|C')+P(C')}=\dfrac{(1)(0.454)}{(1)(0.454+(0.3) (1-0.454)}$
or, $ \approx 0.735$
Thus, we conclude the probability that the victim of a deadly side-impact accident was driving a car is approximately $0.735$.
