## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1106: 101

#### Answer

The fractions of outcomes which is even or greater than 3 is $\frac{2}{3}$.

#### Work Step by Step

We know that the sample space of equally likely outcomes is $S=\left\{ 1,2,3,4,5,6 \right\}$. There are six outcomes in the sample space, so $n\left( S \right)=6$. And the event of getting an even number or a number greater than 3 is $E=\left\{ 2,4,5,6 \right\}$. There are four outcomes in this event, so $n\left( E \right)=4$. Therefore, the probability of rolling an even number or a number greater than 3 is: \begin{align} & P\left( E \right)=\frac{n\left( E \right)}{n\left( S \right)} \\ & =\frac{4}{6} \\ & =\frac{2}{3} \end{align} Thus, the fractions of outcomes which is even or greater than 3 is $\frac{2}{3}$.

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