College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Review Exercises - Page 790: 95

Answer

$\frac{1}{6}$

Work Step by Step

These events are independent so I can find the probability of both happening by multiplying the probability of them happening individually. We know that $probability=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}$ Hence for the first event the probability is: $\frac{2}{6}=\frac{1}{3}$ Hence for the second event the probability is: $\frac{3}{6}=\frac{1}{2}$ Hence the total probability: $\frac{1}{3}\frac{1}{2}=\frac{1}{6}$
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