Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 4 - Probability - 4-6 Counting - Basic Skills and Concepts - Page 181: 19

Answer

a) $\frac{1}{749398}$ b)$\frac{1}{10000}$ c) 10000

Work Step by Step

a) Because the order is not important, we can use the combination rule to determine the number of possibilities: $\frac{41!}{(41-5)!5!}=749398.$ We also know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, therefore $P=\frac{1}{749398}$. b)By using the fundamental counting rule(knowing that there are 10 digits) we can get the number of possibilities: $10\cdot 10\cdot 10\cdot10=10000$. We also know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, therefore $P=\frac{1}{10000}$. c) By using the properties of b), then 1 out of 10000 tickets is a winner, hence if there is no profit the return should be equal to the price of 10000 tickets which is 10000.
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