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

The three applicants make their choices independently and each has three equally likely houses to pick from. Hence the sample space for the ordered triple (choice of person 1, person 2, person 3) has $$3 \times 3 \times 3 = 27$$ equally likely outcomes.

Desired event: all three select the same house.
Case 1: All pick House 1  ⇒  probability $$\left(\tfrac13\right)^3 = \tfrac{1}{27}.$$}
Case 2: All pick House 2  ⇒  probability $$\tfrac{1}{27}.$$}
Case 3: All pick House 3  ⇒  probability $$\tfrac{1}{27}.$$

Because these cases are mutually exclusive, add their probabilities:

$$P(\text{all three choose the same house}) = 3 \times \frac{1}{27} = \frac{1}{9}.$$

Therefore, the required probability is $$\frac{1}{9}.$$

Option B which is: $$\frac{1}{9}$$