Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that three apply for the same house is

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 = \(3\over 3^3\) = \(1\over 9\)

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