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Cosmin has three ping pong balls, and ten cups in front of him. If he throws the ping pong balls into the cups, what is the probability that exactly two balls land in the same cup? (A ping pong ball has an equal probability of landing in any cup.)

 Jun 22, 2022
 #1
avatar+589 
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If there are 10 cups then the chance of a ping pong ball landing in any cup is\(\frac1{10}\)

 

The first ball can go in any cup, but one of the next balls has to land in that same cup, and one has to land in a different cup. So the probability of this is \({\frac1{10}\cdot\frac9{10}=\boxed{\dfrac9{100}}}\)

 Jun 22, 2022
 #2
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I will assume that all the cups are different and all the balls are the same.

 

There are 10C3 ways to choose three cups = 120

There are 10C2 ways to choose 2 cups = 45  

    but either one could have the 2 balls so that is realy 90 ways

there are 10 ways they can all be in the same cup

 

so that makes a total of 220 ways and 90 of those have exactly 2 balls in one cup

 

so     90/220  = 9/22

 

I am not 100% sure it is correct.  I think it is though.

 

When you are given the answer that your teacher/site says is correct please share it with us.

 Jun 22, 2022

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