2010/11 AHSMC Part 1 Problem 7

Problem 7 of 16

A square pigeon coop is divided by interior walls into $9$ square pigeonholes in a $3 \times 3$ configuration. Each of two pigeons chooses a pigeonhole at random, possibly the same one. The probability that they choose two holes on the opposite sides of an interior wall is

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Solution

The pigeons can choose the pigeonholes in $9 \times 9 = 81$ ways. There are $12$ pairs of rooms separated by an interior wall. Since the pigeons can choose these two rooms in $2$ ways, the desired probability is $\frac{2 \times 12}{81} = \frac{8}{27}$ . The answer is (D).

University of Alberta

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