<p>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</p>
Step-by-Step Solution
Key Concept: Each person independently chooses from 3 houses with equal probability 1/3. The total number of ways all three can apply is 3³, and favorable outcomes (all choosing the same house) is exactly 3.
<p><strong>Step 1:</strong> Identify the sample space. Each of the 3 persons independently applies for one of 3 houses. Total number of ways = 3 × 3 × 3 = 3³ = 27</p><p><strong>Step 2:</strong> Count favorable outcomes. All three persons apply for the same house. This can happen in 3 ways: (House 1, House 1, House 1) or (House 2, House 2, House 2) or (House 3, House 3, House 3). So favorable outcomes = 3</p><p><strong>Step 3:</strong> Apply probability formula. P(all three apply for same house) = Favorable outcomes / Total outcomes = 3/27 = 1/9</p><p>∴ Answer: <strong>1/9</strong></p>
Correct Answer: B