Answer
$0.9118$
Work Step by Step
According to Bayes' theorem:
$P(C|D)=\dfrac{P(D|C)P(C)}{P(D|S)P(S)+P(D|C)P(C)+P(D|L)P(L)}~~~~~~~~(1)$
Here, we have
$P(C)=71.1\% =0.711\\ P(S)=5 \% =0.05 \\ P(L)=23.9 \%=0.239\\P(D|C)= 1 \\ P(D|S)=0.371\\ P(D|L)=0.210\\$
Now, we will now use formula (1) and the given data to obtain:
$P(C|D)=\dfrac{P(D|C)P(C)}{P(D|S)P(S)+P(D|C)P(C)+P(D|L)P(L)}=\dfrac{(1)(0.711)}{(0.371)(0.05)+(1)(0.711)+(0.210) (0.239)}$
or, $ \approx 0.9118$
Thus, we conclude the probability that the victim of a deadly side-impact accident was driving a car is approximately $0.9118$.
