Three distinguishable balls are distributed in 3 cells. The probability that all three occupy the same cell, given that atleast two of them are in same cell. (A)`1/7` (B)`1/9` (C)`1/6` (D)None of these

Explanation

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 - Ptext((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`