= Solution
As a rough rare-event calculation, divide the observed elective-group event count by the reported adjusted <odds ratio>:
$$
\boxed{9/2.7\simeq3.3\text{ events}.}
$$
For a homogeneous illustrative risk, the exact odds conversion starts from $p_1=9/1515$ and gives
$$
p_0=\frac{p_1}{2.7(1-p_1)+p_1},\qquad1515p_0\simeq3.35.
$$
The corresponding baseline risk is about $0.22\%$, compared with $1.62\%$ in the observed reference group, roughly seven times lower. Thus \b[taking the adjusted result literally implies exceptionally low background risk among the elective patients].
This is a diagnostic approximation, not an identified mean of <potential outcomes> obtainable from the aggregate table. A conditional <odds ratio> cannot generally be inverted at an aggregate mean: individual background risks are needed for <outcome standardization>. Interpreting the calculation causally also requires <conditional exchangeability>, the <positivity assumption> and a correctly specified outcome model. At very small risks the rare-event approximation makes the estimate of roughly three events useful for checking the plausibility of the adjustment.
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