Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 5 - Section 5.2 - Assess Your Understanding - Applying the Concepts - Page 278: 32a

Answer

P(two or three) $=\frac{2}{13}=0.154$

Work Step by Step

The sample space S = "a standard 52-card deck". So, N(S) = 52. Let E be the event "a two". So, as shown in Figure 9, N(E) = 4 Let F be the event "a three". So, as shown in Figure 9, N(F) = 4 The events "a two" and "a three" are mutually exclusive. So: P(two or three) = P(two) + P(three) = $\frac{N(two)}{N(S)}+\frac{N(three)}{N(S)}=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}=0.154$
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