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 the three apply for the same house, is:

• A. 44751
• B. 44782
• C. 44570
• D. 44601

Answer 1

Answer: (C)

44570

Answer 2

Person 1 has three options to apply. Similarly, person 2 has three options to apply and person 3 has three options to apply. Total cases $= 3^3$ Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3} So probability $= \frac{3}{3^{3}} = \frac{1}{9}.$