1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.