# 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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