Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 12 - Counting and Probability - Section 12.3 Probability - 12.3 Assess Your Understanding - Page 885: 18

Answer

$\dfrac{1}{3}$

Work Step by Step

The experiment involves spinning $\text{Spinner III}$ followed by $\text{Spinner II}$. Thus, the sample space $S$ is: $S=\left\{\text{Forward Yellow}, \text{Forward Green}, \text{Forward Red}, \text{Backward Yellow}, \\ \space \space \space \space \space \space \space \space \space\text{Backward Green}, \text{Backward Red}\right\}$ Note that $n(S)=6$ and that each one has an equal chance of happening (equally-likely outcomes). Let $E_1$ be the event that the outcome is Forward followed by a Green. Then, from the sample space, we have $P(E_1)=\dfrac{1}{6}$. Let $E_2$ be the event that the outcome is Forward followed by a Yellow. Then, $P(E_2)=\dfrac{1}{6}$. Let $E$ = event that a Forward comes out followed by a Green or a Yellow. Then, $P(E)=P(E_1)+P(E_2)=\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{2}{6}=\dfrac{1}{3}$
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