Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Section 7.6 - Bayes' Theorem and Applications - Exercises - Page 519: 16

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 $0.903$.
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