Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 11 - Counting Methods and Probability Theory - 11.7 Events Involving And; Conditional Probability - Exercise Set 11.7 - Page 746: 10

Answer

$\frac{1}{27}$

Work Step by Step

If A, B and C are independent events, then P(E) = P(A)*P(B)*P(C) E: green every time A: green B: green C: green P(A) = $\frac{2}{6}$ P(B) = $\frac{2}{6}$ P(C) = $\frac{2}{6}$ All of the events are independent of each other, so it follows that: P(E) = $\frac{2}{6}$ .$\frac{2}{6}$ . $\frac{2}{6}$ = $\frac{1}{27}$
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