## Stats: Data and Models (3rd Edition)

$P(ace|suit)=\frac{P(\text{ace and suit)}}{P(suit)}=\frac{1/52}{13/52}=\frac{1}{13}\approx0.0769=7.69\%$ $P(ace)=\frac{4}{52}=\frac{1}{13}\approx0.0769=7.69\%$ Thus, yes, because both probabilities are the same.