= Solution
Pearson's statistic is
$$
\boxed{X^2=\sum_{i=1}^4\sum_{j=1}^3\frac{(O_{ij}-E_{ij})^2}{E_{ij}}.}
$$
For <independent> observations under the <independence> <null hypothesis>, its large-sample null distribution is approximately $\chi^2$ with $(4-1)(3-1)=6$ <statistical degrees of freedom>. The fitted row and column probabilities account for the reduction in <statistical degrees of freedom>. All <expected counts> exceed five, so the usual approximation has adequate expected cell sizes here. Reject <independence> when the statistic exceeds the corresponding upper-tail <critical value>.
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