College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 10 - Section 10.3 - Probability - 10.3 Assess Your Understanding - Page 706: 68

Answer

See below.

Work Step by Step

Since the events are independent: a) $P(\text{at most 2})=P(0)+P(1)+P(2)=0.1+0.15+0.2=0.45$ b) $P(\text{at least 2})=P(2)+P(3)+P(4 \ or \ more)=0.2+0.24+0.31=0.75$ c) $P(\text{at least 1})=P(1)+P(2)+P(3)+P(4 \ or \ more)=0.15+0.2+0.24+0.31=0.9$
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