We distribute the 10 balls into 6 bins. If we randomly distribute the balls into the bins, what is the probability that all of the balls end up in the same bin? Is it just: (${10 \choose 1}*6)/ 6^{10}$
All balls fall into 1 bin
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probability
combinatorics
balls-in-bins
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0Where did the $\binom{10}{1}$ come from? – 2017-02-11
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010 balls and choose 1 bin to put them all in, since there are 6 bins we can do this 6 different ways 10*6/6^10 – 2017-02-11
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2That would be $\binom 6 1 1^{10}/6^{10}$, since we are choosing 1 from 6 bins, *not* 1 from 10 balls; which simplifies to $1/6^9$. – 2017-02-11
1 Answers
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The first ball can go in any of the bins ... but after that all other 9 need to go in the same bin, so the chances of that are $(\frac{1}{6})^9$
The fact that you have 10 balls does not multiply this by 10.