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.)
+592 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 \({\frac1{10}\cdot\frac9{10}=\boxed{\dfrac9{100}}}\)
+118687 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.
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