= Solution
Treat the counts as independent <binomial distributions>, $R_i\sim\operatorname{Bin}(n_i,\pi_i)$, conditional on the known numbers of operations. The <null hypothesis> is that the probabilities $\pi_i$ are all equal, with no assumed numerical value. Centre 4 has no operations in this period and hence contributes <likelihood> one, supplying no information about its mortality <probability>. The <empty groups in a binomial homogeneity test> principle leaves \b[11 contributing centres and 10 test degrees of freedom].
Under the common-probability model the <maximum-likelihood estimator> is
$$
\widehat\pi=\frac{\sum_i r_i}{\sum_i n_i}=\frac{283}{1353}=0.209165.
$$
Use the two-column <contingency table> of deaths and survivors, deleting its all-zero row. The <Pearson chi-squared test of homogeneity> has statistic
$$
X^2=\sum_{i:n_i>0}\left\{\frac{(r_i-n_i\widehat\pi)^2}{n_i\widehat\pi}+\frac{((n_i-r_i)-n_i(1-\widehat\pi))^2}{n_i(1-\widehat\pi)}\right\}
=\sum_{i:n_i>0}\frac{(r_i-n_i\widehat\pi)^2}{n_i\widehat\pi(1-\widehat\pi)}.
$$
Under the null, the large-sample reference distribution is $\chi^2_{10}$. The expected counts are adequate for this approximation here, including at the smallest nonempty centre. Calculating gives
$$
\boxed{X^2=39.3923,\qquad p\approx2.17\times10^{-5}.}
$$
An alternative is the <likelihood-ratio test> against separate centre probabilities, with <binomial deviance>
$$
D=2\sum_{i:n_i>0}\left\{r_i\log\frac{r_i}{n_i\widehat\pi}+(n_i-r_i)\log\frac{n_i-r_i}{n_i(1-\widehat\pi)}\right\}.
$$
The unrestricted probabilities are $r_i/n_i$; subtracting one fitted common <probability> leaves the same 10 degrees of freedom. Here $D=39.5033$ and $p\approx2.07\times10^{-5}$. Thus \b[there is strong evidence against a common mortality <probability>]. This omnibus test does not identify Centre 1 as the sole source of the heterogeneity. If its comparison with the remaining centres is the primary scientific question, specify a separate planned one-versus-rest comparison; inspection of many pairwise comparisons calls for consideration of <multiple testing>.
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