Question:
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 three apply for the same house is:
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 three apply for the same house is:
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Alternatively: the first person can choose any house (prob = 1). The second and third persons must choose that same house, each with probability $1/3$. Total probability $= 1 \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.
Updated On: May 31, 2026
- $\frac{2}{9}$
- $\frac{1}{9}$
- $\frac{4}{9}$
- $\frac{1}{27}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The probability is calculated by dividing the number of favorable ways by the total possible ways of applying for the houses.
Step 2: Meaning
Each of the 3 persons has 3 choices of houses. We want to find the probability that they all select the exact same house.
Step 3: Analysis
Total possible ways the 3 persons can apply for the houses: \[ n(S) = 3 \times 3 \times 3 = 3^3 = 27 \] Favorable ways where all three apply for the same house: \[ n(E) = 3 \quad (\text{either all choose House 1, all choose House 2, or all choose House 3}) \] Calculating the probability: \[ P(E) = \frac{n(E)}{n(S)} = \frac{3}{27} = \frac{1}{9} \]
Step 4: Conclusion
The probability that all three persons apply for the same house is $\frac{1}{9}$. Final Answer: (B)
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