Finite Math and Applied Calculus (6th Edition)

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

Chapter 7 - Section 7.3 - Probability and Probability Models - Exercises - Page 483: 21

Answer

See the picture. $P(\lt4)=\frac{4}{9}$

Work Step by Step

$P(2)=P(4)=P(6)=x$ $P(1)=P(3)=P(5)=y$ Such that $x=2y$ $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1$ $3x+3y=1$ $3(2y)+3y=1$ $9y=1$ $y=\frac{1}{9}$ $x=2y$ $x=\frac{2}{9}$ $P(2)=P(4)=P(6)=\frac{2}{9}$ $P(1)=P(3)=P(5)=\frac{1}{9}$ $P(\lt4)=P(1)+P(2)+P(3)=\frac{1}{9}+\frac{2}{9}+\frac{1}{9}=\frac{4}{9}$
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