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Let $X$ and $Y$ be independent random variables with binomial distributions $\mathrm{Bin}(n,p)$ and $\mathrm{Bin}(m,q)$. Conditional on $X+Y=k$ having positive probability, the possible values of $X$ satisfy $\max (0,k-m)\le j\le \min (n,k)$. Their probabilities arewhere the denominator sums over that feasible integer range. The formula follows by multiplying independent binomial probabilities and dividing by the probability of the disjoint union of all feasible count pairs.

For $p,q\in (0,1)$, common factors cancel and the weights are proportional to $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{m}{k-j}\right){\theta }^j$, where $\theta =p(1-q)/[q(1-p)]$. If $p=q$, this reduces to the hypergeometric distribution, with probabilities $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{m}{k-j}\right)/\left(\genfrac{}{}{0ex}{}{n+m}{k}\right)$. At parameter endpoints, use the original probability weights and check that the conditioning event has positive probability before dividing.

The association remains positive within each income group, although its estimated size is smaller in the higher-income groups. The two extreme trust categories give the following conditional vaccination percentages and differences:

In the lowest-income group, the intervening percentages are 42.86%, 44.12% and 82.54%, so the increase is monotone. Its Pearson chi-squared statistic is approximately 62.520 on four statistical degrees of freedom, with the printed asymptotic $p\approx 8.56\times 10^{-13}$. Cramér's V is approximately 0.441, Goodman and Kruskal's gamma is approximately 0.626 and the column-dependent Somers' D is approximately 0.259. The column-dependent Goodman and Kruskal's lambda is approximately 0.158: trust improves the modal prediction more here than in the pooled table.

In the middle-income group, the intervening vaccination percentages are 61.54%, 81.82% and 92.04%. The printed Pearson chi-squared statistic is approximately 24.033 on four statistical degrees of freedom, with asymptotic $p\approx 7.87\times 10^{-5}$. Cramér's V is approximately 0.298, Goodman and Kruskal's gamma approximately 0.454 and column-dependent Somers' D approximately 0.100. Column-dependent Goodman and Kruskal's lambda is zero: vaccinated respondents are the majority in every row, so the modal prediction never changes. A zero lambda does not mean independence.

In the highest-income group, the intermediate vaccination percentages are 71.43%, 76.92% and 94.55%; the final category falls slightly to 91.14%. Thus the observed pattern is broadly positive rather than strictly monotone. The printed Pearson chi-squared statistic is approximately 8.199 on four statistical degrees of freedom, with asymptotic $p\approx 0.0846$. It does not reject independence at 5%. The corresponding Cramér's V is approximately 0.229, Goodman and Kruskal's gamma approximately 0.281 and column-dependent Somers' D approximately 0.058. The column-dependent Goodman and Kruskal's lambda is again zero.

Small counts limit the asymptotic tests, especially in the middle- and highest-income tables. Their minimum expected counts are approximately 0.644 and 0.306, with respectively 30% and 40% of cells below five. The lowest-income table has one expected count below five, with minimum approximately 3.315. These diagnostics concern expected counts under independence, not just the observed row sizes.

The supplied counts are sufficient to make an exact conditional Pearson test for a binary contingency table, without needing individual records or assuming a chi-squared distribution approximation. Conditioning on the row totals and total unvaccinated count removes the common unknown binomial probability. For each admissible table, sum its multivariate hypergeometric distribution probability when its Pearson chi-squared statistic is at least the observed one. Complete enumeration gives:

These are new exact conditional calculations, rather than the paper's printed asymptotic p-values. They confirm evidence of association in the first two groups and nonrejection at 5% in the third, while showing the numerical effect of sparse cells. The tail is ordered by the Pearson chi-squared statistic; an exact test with a different ordering need not have the same p-value. Fixed-margin Monte Carlo tests approximate this same Pearson-ordered tail when enumeration is impractical.^[ref]

The original count-based figure below includes pointwise approximate 95% Wilson intervals. The very wide intervals for sparsely populated trust categories make the uncertainty visible; they are not simultaneous intervals or a formal test of differences between income groups.

In the elaboration model, the positive direction is replicated after conditioning on income: income does not fully explain away the observed trust–vaccination relationship. The decreasing estimated contrasts and ordered coefficients suggest possible specification by income, or effect modification on an explicitly chosen scale. They do not establish it. Different p-values can arise from different sample sizes and standard errors; significant and nonsignificant results need not differ significantly. A formal statistical interaction test or a confidence interval for differences between comparable contrasts would address that question.^[ref]

Nor do these conditional tables prove that income is a causal confounder or causal mediator. Those labels require assumptions about causal order, relevant confounding and how variables were measured. This is an observational study of recorded vaccination status, not a randomized change in trust or a test of vaccine effectiveness.

The number $R$ of successes has a binomial distribution. Each arrangement of $r$ successes and $n-r$ failures has probability $p^r{q}^{n-r}$ by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing $R$ as a sum of independent Bernoulli random variables gives its expectation and variance: $\mathbb{E}R=np$, $\mathrm{Var}(R)=npq$. Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts $np$ and $nq$ are sufficiently large; large $n$ alone is insufficient if $p$ is extremely close to zero or one.

For the guessing student, $n=60$, $p=1/4$, $q=3/4$. Therefore the expected number of correct answers is $\boxed{15}$, and ${\sigma }^2=45/4$, $\sigma =3\sqrt{5}/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\le X\le 17.5$ after the continuity correction. Standardizing the normal distribution givesHere $(5/2)/(3\sqrt{5}/2)=\sqrt{5}/3$, and the last step uses the supplied bound. The approximate probability is $0.54394$. The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately $0.54370$, also exceeding one half.