As each ball can be placed in a cell in three different ways, all the three distinct balls can be distributed in three cells in 3*3*3 = 27 ways.

Let A, B are the events defined as:

A : all balls are in the same cell

B : atleast two balls are in the same call

Now, all balls can be placed in the same cell in 3 ways

So, P(A) = 3/27 = 1/9

and P(B) = 1 - P(Balls are placed in different cells)

=> P(B) = 1 - 3!/27

=> P(B) = 1 - 6/27

=> P(B) = (27 - 6)/27

=> P(B) = 21/27

=> P(B) = 7/9

Since A ⊂ B, A ∩ B = A

So, P(A∩B) = P(A) = 1/9

Required probability = P(A/B) = P(A∩B)/P(A) = (1/9)/(7/9) = 1/7