## Solution :

\(\therefore\) Total number of cases = \(3^3\)

Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3)

\(\therefore\) Required Probability = \(3\over 3^3\) = \(1\over 9\)

\(\therefore\) Total number of cases = \(3^3\)

Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3)

\(\therefore\) Required Probability = \(3\over 3^3\) = \(1\over 9\)