= Solution
Using the reference event rate as the background-risk benchmark gives
$$
\boxed{1515\frac{1215}{75057}\simeq24.5,\quad\text{about 25 events}.}
$$
This is the expected count if the elective group had the reference group's baseline risk and the procedure added no effect. It is substantially above the observed count, illustrating the importance of the background-risk comparison.
Similarity of background factors alone does not fix the elective group's event risk if the procedure itself changes it. If one additionally retained the reported common adjusted <odds ratio> $R=2.7$ and used $p_0=1215/75057$ as the common baseline risk, then conversion from odds to <probability> would give
$$
p_1=\frac{Rp_0}{1-p_0+Rp_0},\qquad1515p_1\simeq64.4.
$$
These are different assumptions: \b[about 25 is the baseline/no-effect benchmark; about 64 includes an odds multiplier of 2.7].
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