# Thread: Probabilty, Putting balls in boxes

1. ## Probabilty, Putting balls in boxes

I have 100 balls that are numbered and 100 boxes that are numbered. Each box can only contain one ball. The balls are randomly placed in the boxes. What is the probability that at least one ball ends up in the same numbered box?

I know that the number of ways I can choose the 100 balls is 100!

I'm not really sure where to go next. Would it be easier to look at the event that none of the balls are in the same numbered box?

2. The number of ways for no ball to be is its own box is $D(100) = \left( {100!} \right)\sum\limits_{k = 0}^{100} {\frac{{\left( { - 1} \right)^k }} {{k!}}} \approx \frac{{100!}}{e}$.

3. event A: at least one ball in same numbered box
event B: all of the balls in the same numbered box
event C:none of the balls in the same numbered box equal

Does P(C)=1-P(A) or 1-P(B)?

I know also that P(B) is 1/(100!) since there are 100! ways to choose the balls and only one of the ways puts them in order.

Where does that summation come from? What is that function D called?

4. Originally Posted by ebot
event A: at least one ball in same numbered box
event B: all of the balls in the same numbered box
event C:none of the balls in the same numbered box equal

Does P(C)=1-P(A) or 1-P(B)?

I know also that P(B) is 1/(100!) since there are 100! ways to choose the balls and only one of the ways puts them in order.

Where does that summation come from? What is that function D called?
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