Finite Math and Applied Calculus (6th Edition)

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

Chapter 7 - Section 7.4 - Probability and Counting Techniques - Exercises - Page 493: 15

Answer

$\displaystyle \frac{1}{5}$

Work Step by Step

$P(E)=\displaystyle \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$ $n(S)=C(10,2)=45$ E=( 1 of PFE) AND ( 1 of the other 9) $n(E)=C(1,1)\cdot C(9,1)=9$ $P(E)=\displaystyle \frac{9}{45}=\frac{1}{5}$
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