= Solution
Test \b[homogeneity of the four category <probabilities>] among the three samples. The <null hypothesis> is $p_{iA}=p_{iB}=p_{iC}$ for every category $i$; the alternative allows at least one difference. This is a <Pearson chi-squared test of homogeneity>, assuming independently sampled grains with consistent classification.
The row totals are $(195,52,64,16)$, the sample totals $(131,135,61)$ and the grand total $N=327$. Under the <null hypothesis>, $E_{ij}=O_{i+}O_{+j}/N$, giving the <expected value> table
$$
\begin{array}{c|rrr}
& A&B&C\\\hline
\text{non-undulatory}&78.1193&80.5046&36.3761\\
\text{undulatory}&20.8318&21.4679&9.7003\\
\text{polycrystalline, 2–3}&25.6391&26.4220&11.9388\\
\text{polycrystalline, greater than 3}&6.4098&6.6055&2.9847
\end{array}
$$
For example, $E_{\mathrm{undulatory},C}=52(61)/327=9.7003$, so that cell contributes $(4-9.7003)^2/9.7003=3.34974$. The complete cell contributions are
$$
\frac{(O-E)^2}{E}=
\begin{pmatrix}
0.217211&0.003049&0.361014\\
0.481834&0.298659&3.349738\\
0.217388&0.222018&0.000313\\
0.310072&0.055505&1.360734
\end{pmatrix},
$$
so \b[$X^2=6.87754$]. There are $(4-1)(3-1)=6$ <statistical degrees of freedom>. The supplied $5\%$ critical value is $12.592$, and $6.87754<12.592$, so \b[do not reject homogeneity at $\alpha=0.05$]. The approximate upper-tail <p-value> is $0.3323$.
One of twelve expected cells is below five, with none below one; the asymptotic approximation is broadly defensible by a common expected-count guideline but is not exact. An exact/conditional simulation could check it, especially if grain counts are clustered. Do not pool categories solely to obtain a desired result. The conclusion is that these counts do not demonstrate different category proportions; \b[they do not prove that the rocks share one <geological formation>]. Different formations can share such proportions, and real grains within one rock can fail the independent-sampling assumption.
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