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 518: 1

Answer

$P(B|A)=\frac{.8\times.2}{.8\times.2+.3\times .8}=.4$

Work Step by Step

Accordint to Bayes' theorem: $P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A')P(A')}$ Here, we have $P(A|B)= .8$ $P(B)= .2$ $P(A|B')= .3$ Be aware, that A and B are in the definition of the theorem. We know, that $P(B')=1-P(B)=1-.2=.8$ We can substitute into the definition: $P(B|A)=\frac{.8\times.2}{.8\times.2+.3\times .8}=.4$
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