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:

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) 7/9 (B) 8/9 (C) 1/9 (D) 2/9

Tardigrade Admin · Accepted Answer

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} .

Q. 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:

7/9

8/9

1/9

2/9

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 =33 Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3} So probability =333​=91​.

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 $= \frac{3}{3 ^{3}} = \frac{1}{9} .$