So, number of favorable selections:
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- There are 4 rows, choose 3 distinct rows: \( \binom{4}{3} = 4 \).
- Each chosen row has 3 columns, so for each row, pick 1 position: \( 3 \times 3 \times 3 = 27 \).
- But this counts selections per row order — since we are selecting unordered triples, we must divide by \( 3! = 6 \) only if order doesn’t matter. However, since we are selecting positions and the grid is fixed, each position is unique, and choosing one per row from 3 distinct rows gives \( 3^3 = 27 \) per set of 3 rows.
- \binom{4}{3} \times 3^3 = 4 \times 27 = 108
- Thus, probability:
- P = \frac{108}{220} = \frac{27}{55}
