## Statistics: Informed Decisions Using Data (4th Edition)

$P(flushing~|~viagra)=\frac{73}{734}\approx0.0995\gt0.05$. So, we can conclude that "a subject from the Viagra group reported experiencing flushing" is not an unusual event.
The Viagra group sample has 734 subjects. So, $N(S)=734$. Event = "subject from the Viagra group reported experiencing flushing". $N(E)=73$. Using the Classical Method: $P(E)=P(flushing~|~viagra)=\frac{N(E)}{N(S)}=\frac{73}{734}\approx0.0995\gt0.05$