 Solution

Under independence, the cell probability is the product of its row and column probabilities. Estimate these by the sample proportions. If $R_i,C_j$ are row and column totals and $N$ is the grand total, the fitted expected count isHere the row totals are $123,76,163,90$, the column totals $304,65,83$, and $N=452$. For example $E_{11}=123\cdot 304/452=82.72566\dots$, rounding to 82.73 as printed. The other entries follow identically.