Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 12 - Data Analysis and Probability - Concept Byte - Conditional Probability - Page 771: 7


Yes. There are nine ways for there to be an even number on cube one and an odd number on cube two. The conditional probability is the same for both events.

Work Step by Step

number cubes are numbered $1, 2, 3, 4, 5,$ and $6$ (three even numbers and three odd numbers) P(even on cube 1 | odd on cube 2) = $1/4$ P(odd on cube 2 | even on cube 1) = $1/4$
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