= Solution
Let $X_j$ be the number of failures in mix $j$. The total observed failures are $32+2(12)+3(2)+4(1)=66$ among $600$ blocks. Equal marginal failure <probabilities> give $E(X_j)=6p$, regardless of dependence within a mix. Thus the <unbiased estimator> based on <grouped Bernoulli failure counts> is
$$
\boxed{\widehat p=\frac{\sum_{j=1}^{100}X_j}{600}=0.11.}
$$
For a conventional binomial <confidence interval>, additionally assume independent block outcomes. Then the total $S$ has <binomial distribution> $\operatorname{Bin}(600,p)$ and the estimated <standard error> is $\sqrt{\widehat p(1-\widehat p)/600}=0.01277$. The <normal approximation> gives the approximate 95 percent interval
$$
\boxed{0.11\pm1.96(0.01277)\simeq(0.0850,0.1350).}
$$
One can instead obtain an exact, generally conservative binomial interval by solving $P_{p_L}(S\ge66)=0.025$ and $P_{p_U}(S\le66)=0.025$ for the two endpoints. These equations invert binomial tail tests.
Equal marginal <probabilities> alone do not imply independent blocks. If the mixes are independent but blocks within a mix can be dependent, use the <sample variance> of the 100 counts. Here $\sum X_j^2=114$ and $\overline X=0.66$, so $s_X^2=(114-100(0.66)^2)/99=0.7115$. The estimated <variance> of $\widehat p=\overline X/6$ is $s_X^2/(100\times36)$; its <standard error> is $0.01406$, giving the large-sample interval approximately $(0.0824,0.1376)$. If even between-mix independence is absent, no stated model justifies either numerical coverage claim. The estimate remains unbiased, but the uncertainty calculation needs its sampling assumptions.
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