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: (${10 \choose 1}*6)/ 6^{10}$

  • $\begingroup$ Where did the $\binom{10}{1}$ come from? $\endgroup$ – David K Feb 11 '17 at 20:18
  • $\begingroup$ 10 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 $\endgroup$ – user1775500 Feb 11 '17 at 20:58
  • 2
    $\begingroup$ That 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 $1/6^9$. $\endgroup$ – Graham Kemp Feb 11 '17 at 23:00

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.


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