Thre 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.
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hey dude....
here is ur answer....!!!
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
here is ur answer....!!!
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
swarupsau:
Thank you
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