I have 4 urns and 4 balls.

With each turn, I randomly place 1 ball in 1 urn.

I take a total of 4 turns.

What is the probability of exactly 0 empty urns?

P(0 empty urns) = P(all 4 balls are placed in 1 urn each) =

P(1st ball placed anywhere).P(2nd ball placed anywhere of the 3 urns left) ....P(and so on)=

$\displaystyle {4 \over 4}.{3 \over 4}. {2 \over 4}.{1 \over 4} = {24 \over 256}$

What is the probability of exactly 4 empty urns?

Impossible. Probability is ZERO.

What is the probability of exactly 3 empty urns?

P(3 empty urns) = P(all 4 balls are placed in the same urn) =

$\displaystyle {4 \over 4}.{1 \over 4}. {1 \over 4}.{1 \over 4} = {4 \over 256}$

What is the probability of exactly 1 empty urn?

P(1 empty urn) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns left).P(4th ball in any of the 3 urns originally chosen)=

$\displaystyle {4 \over 4}.{3 \over 4}. {2 \over 4}.{3 \over 4} = {72 \over 256}$

What is the probability of 2 empty urns?

P(2 empty urns) = P(1st ball anywhere).P(2nd ball in any of the 3 urns left).P(3rd ball in any of the 2 urns originally chosen).P(4th ball in any of the 2 urns originally chosen)=

$\displaystyle {4 \over 4}.{3 \over 4}. {2 \over 4}.{2 \over 4} = {48 \over 256}$

Ok. Now comes the problem!

I know that they should all add up to the grand probability of 1.

But they don't. Where did I go wrong???

Thanks for reading!