AIEEE 2005 Math Solved Paper

Person 1 st has three options to apply.
Similarly, person 2nd has three options t apply and person 3rd 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 =‌

3

33

=‌

1

9

.