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 748: 74

Answer

See below.

Work Step by Step

If the events are independent, $P(\text{A and B})=P(A)P(B)$. E.g. if we draw number from $1$ to $5$ from a hat with replacement, then the probability of drawing a $3$ and then a $4$ is: $\frac{1}{5}\frac{1}{5}=\frac{1}{25}$
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