Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 13 - Counting and Probability - 13.3 Probability - 13.3 Asses Your Understanding - Page 865: 14

Answer

We have the outcomes: $H1H,H2H,H3H,H4H,H5H,H6H,T1H,T2H,T3H,T4H,T5H,T6H,H1T,H2T,H3T,H4T,H5T,H6T,T1T,T2T,T3T,T4T,T5T,T6T.$ These are $24$ equally likely cases, hence all of their probability is $\frac{1}{24}$.

Work Step by Step

We have the outcomes: $H1H,H2H,H3H,H4H,H5H,H6H,T1H,T2H,T3H,T4H,T5H,T6H,H1T,H2T,H3T,H4T,H5T,H6T,T1T,T2T,T3T,T4T,T5T,T6T.$ These are $24$ equally likely cases, hence all of their probability is $\frac{1}{24}$.
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