Answer
Yes.
Work Step by Step
$P(ace|suit)=\frac{P(\text{ace and suit)}}{P(suit)}=\frac{1/52}{13/52}=\frac{1}{13}\approx0.0769=7.69\%$
$P(ace)=\frac{4}{52}=\frac{1}{13}\approx0.0769=7.69\%$
Thus, yes, because both probabilities are the same.
Stats: Data and Models (3rd Edition)
by De Vaux, Richard D.; Velleman, Paul D.; Bock, David E.
Published by
Pearson
ISBN 10:
0321692551
ISBN 13:
978-0-32169-255-9
Chapter 15 - Probability Rules! - Exercises - Page 377: 23
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