= Solution
The <odds ratio> compares event odds, not event probabilities. Let group 1 denote the elective group and group 0 the reference group. Their fitted event <probabilities> are $\widehat p_1=9/1515\simeq0.00594$ and $\widehat p_0=1215/75057\simeq0.01619$. Thus
$$
\boxed{\widehat{\mathrm{OR}}_{\rm crude}
=\frac{9/(1515-9)}{1215/(75057-1215)}
=\frac{9\cdot73842}{1506\cdot1215}\simeq0.363.}
$$
The <risk ratio> is similarly about $0.367$. \b[The unadjusted observations show a lower adverse-event frequency in the elective group], rather than an increase. This is an association, not evidence that choosing the procedure prevents adverse outcomes. The direction is opposite to the reported adjusted <odds ratio> above one; the two analyses compare different populations or different conditional risks.
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