NIMCET 2016 Question Paper with solutions

Since, each person can apply for any house.
Therefore, total number of ways in which three person can apply for three houses
$3\times 3\times 3$
$=27$ ways
As given that three person can apply for house 1 or house 2 or house 3 among three house.
So, three person can apply for the same house in 3 ways.
∴ Favourable number of ways = 3
Hence, required probability

$\frac{3}{27}=\frac{1}{9}$