Answer
$P(Bryce~|~Gourmet)=\frac{3}{88}\approx0.03409$
It is an unusual event.
Work Step by Step
The sample space: 1009 cases. So, $N(S)=1009$
According to the marginal distribution (see page 235) of the first column: $N(Gourmet)=264$
According to the cell in the third row, first column: $N(Bryce~and~Gourmet)=9$
Using the Conditional Probability Rule (page 288):
$P(Bryce~|~Gourmet)=\frac{N(Bryce~and~Gourmet)}{N(Gourmet)}=\frac{9}{264}=\frac{3}{88}\approx0.03409$
$P(Bryce~|~Gourmet)\lt0.05$. So, it is an unusual event.
