## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Review Exercises - Page 1126: 95

#### Answer

The probability is $\frac{1}{6}$.

#### Work Step by Step

We know that there are six outcomes, so $\text{n}\left( \text{S} \right)=\text{ 6}$. There are two outcomes in the green stopping event, so $\text{n}{{\left( \text{E} \right)}_{\text{green}}}=\text{ 2}$. And the probability of stopping on red is given below, $\text{P}{{\left( \text{E} \right)}_{\text{green}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{green}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{2}{6}\text{ }$ And the event of stopping on a number less than 4 can be represented by ${{\left( \text{E} \right)}_{\text{less than 4}}}\text{ }=\text{ }\left\{ 1,2,3 \right\}$ There are three outcomes in this event, so $\text{n}{{\left( \text{E} \right)}_{\text{less than 4}}}=\text{ 3}$. And the probability of stopping on less than $4$ is given below $\text{P}{{\left( \text{E} \right)}_{\text{less than 4}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{less than 4}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{3}{6}\text{ }$ Thus, the probability of stopping on green on first spin and a number less than $4$ on second spin is \begin{align} & \text{P}\left( \text{red on first spin and less than 4 on second spin} \right)=\text{P}\left( \text{red} \right)\times \text{P}\left( \text{less than 4} \right) \\ & =\frac{2}{6}\times \frac{3}{6} \\ & =\frac{6}{36} \\ & =\frac{1}{6} \end{align}

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