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: 21

Answer

$\frac{1}{64}$

Work Step by Step

P(E) = P(A)*P(B) Assume that we are flipping a coin in a fair manner. Find the probability of some series of events, E, happening. Let A be the first event, B be the second event, and so on. We consider the chances of getting heads six times in a row: P(A) =P(B)=P(C)=P(D)=P(E)=P(F)= $\frac{1}{2}$ P(E) = P(A)*P(B)*P(C)*P(D)*P(E)*P(F) P(E) = $\frac{1}{2^{6}}$ = $\frac{1}{64}$
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