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 747: 30

Answer

0.001

Work Step by Step

If A and B are independent events, then P(E) = P(A)*P(B) A: frequent hangovers B: frequent hangovers C: frequent hangovers P(A) = 0.10 P(B) =0.10 P(C)= 0.10 P(E) = P(A)$\times$P(B)$\times$P(C) P(E) =0.10 x 0.10 x 0.10 =0.001 Thus, the probability that three randomly selected adults will all suffer from frequent hangovers is .001.
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