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: 13

Answer

We have the outcomes: $HH1,HH2,HH3,HH4,HH5,HH6,HT1,HT2,HT3,HT4,HT5,HT6,TH1,TH2,TH3,TH4,TH5.TH6,TT1,TT2,TT3,TT4,TT5,TT6.$ These are $24$ equally likely cases, hence all of their probability is $\frac{1}{24}$.

Work Step by Step

We have the outcomes: $HH1,HH2,HH3,HH4,HH5,HH6,HT1,HT2,HT3,HT4,HT5,HT6,TH1,TH2,TH3,TH4,TH5.TH6,TT1,TT2,TT3,TT4,TT5,TT6.$ These are $24$ equally likely cases, hence all of their probability is $\frac{1}{24}$.
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