Question

Three distinguishable ball distributed in three cells. Find the conditional probability that all the three occupy the same cell, given that at least two of them are in the same cell.

Answer 1

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