Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Review - Review Exercises - Page 536: 1

Answer

The total number of outcomes is equal to 8 $(n(S) = 8)$ The elements of the event $E$ are: $E = \{ HHT, HTH, THH, HTT, TTH, THT, TTT\}$ And the probability of $E$ is equal to $\frac 7 8$

Work Step by Step

Three coins are tossed. 1. A coin toss can result in Head or Tails, both outcomes with the same probability of occurring, so: $S = \{HHH, HHT, HTH, THH, HTT, TTH, THT, TTT\}$ Therefore: $n(S) = 8$, which is the total number of outcomes. 2. The set: "E" will have all the outcomes with one or more tails: $E = \{ HHT, HTH, THH, HTT, TTH, THT, TTT\}$ $n(E) = 7$ 3. Calculate the probability: $P(E) = \frac{n(E)}{n(S)} = \frac 7 8$
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