= Solution
Under the null hypothesis the three outcome probabilities are the same in the two independent samples. The pooled row totals are $50,20,30$, and both column totals are $50$, out of a total of $100$. Thus the expected counts in each column are $25,10,15$. These exceed the usual small-count threshold for the asymptotic <Pearson chi-squared test of homogeneity>.
The <Pearson chi-squared statistic for contingency tables> is
$$
X^2=2\left(\frac{(28-25)^2}{25}+\frac{(4-10)^2}{10}+\frac{(18-15)^2}{15}\right)=\boxed{9.12}.
$$
Under the null, its asymptotic <chi-squared distribution> has $(3-1)(2-1)=2$ <degrees of freedom>. Equivalently, the two independent three-category samples have four probability parameters unrestricted and two under the common-probability null. The $1\%$ upper-tail critical value is $9.21$.
Since $9.12<9.21$, \b[do not reject the equal-effect hypothesis at the 1% level]. For two <degrees of freedom>, the approximate <p-value> is $e^{-9.12/2}\simeq0.01046$, just above $0.01$. This is a decision at the specified <significance level>, not evidence that the two outcome distributions have been proved identical.
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