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 - Page 747: 40

Answer

$\frac{2}{21}$

Work Step by Step

We know that $P(\text{A and B})=P(A)P(B|A)$ because the events are dependent. Hence $P(\text{A and B})=\frac{4}{15}\frac{5}{14}=\frac{2}{21}$
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