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There are $6$ urns indexed from $1$ to $6$. We place $15$ balls one by one randomly into the urns.

What is the probability that the first urn stays empty?

I found $\frac{19\choose4}{20\choose5}$ because there are ${15+6-1\choose 6-1}$ way to distribute the balls in total and ${19\choose4}$ ways to distribute the $15$ balls into $5$ urns ($2$ through $6$).

I have trouble with the following questions:

1) What is the probability that the urns $1,2,3,4$ contain $3$ balls, the $5$th urn contains $2$ balls and the $6$th urn contains $1$ balls?

2) What is the probability that $4$ urns contain $3$ balls each, one urn contains $2$ balls and $1$ contains 1 ball?

For 1) I thought I could use the multinomial coefficient to obtain $\frac{15\choose3,3,3,3,2,1}{20\choose5}$ but this is not a probability (by a huge margin).

For 2) it should be whatever we got for 1) multiplied by $6!$ because that's the number of permutations of the $6$ urns

Edit Maybe this is correct for 1):

${15\choose 3,3,3,3,2,1}({1\over6})^3({1\over6})^3({1\over6})^3({1\over6})^3({1\over6})^2({1\over6})^1={15\choose 3,3,3,3,2,1}({1\over6})^{15}$

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  • $\begingroup$ Your first calculation is incorrect because those ways of populating the urns are not equi-probable. The probability that the first urn is empty is $\left(\frac 56\right)^{15}$. $\endgroup$ – lulu Jan 29 '19 at 11:56
  • $\begingroup$ But I'm just counting the number of ways that it works and dividing by the total number of ways, so I must be counting wrong. $\endgroup$ – H. Walter Jan 29 '19 at 11:57
  • $\begingroup$ No, you counted right. But the distributions are not equiprobable so you can't use the count to get the probability. $\endgroup$ – lulu Jan 29 '19 at 11:59
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Your first calculation is incorrect because those ways of populating the urns are not equi-probable. The probability that any specified ball misses the first urn is $\frac 56$ so the probability that the first urn is empty is $\left(\frac 56\right)^{15}$.

To see the difference between the two probability distributions, just consider the $2-$ball, $2-$ urn case. There is a $\frac 14$ chance that both balls go in the first urn, and a $\frac 14$ chance they both go in the second. But there is a $\frac 12$ chance that they go into different urns.

To do the first question, say, there are $\binom {15}3$ ways to pick the balls that go into the first urn, then $\binom {12}3$ ways to pick the balls that go into the second urn, and so on. We come to the result $$\binom {15}3\times \binom {12}3\times \binom {9}3\times \binom {6}3\times \binom 32\times \left(\frac 16\right)^{15}\approx .0011$$

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  • $\begingroup$ Yes I just realized and edited my question, and our results are the same. Do you agree about 2)? $\endgroup$ – H. Walter Jan 29 '19 at 12:07
  • $\begingroup$ No I don't. There are $\binom 64$ ways to choose the $3-$ball urns., and then $\binom 21$ ways to choose the $2-$ ball urn. So you need to take the product of those as your factor. $\endgroup$ – lulu Jan 29 '19 at 12:10
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For 2)

First we must decide which urns contain 3 balls, which 2 balls, etc.

$A=\binom {6}4\binom {2}1\binom {1}1$.

And answer for 2) is answer from 1) multiplied by $A$.

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There's a 1:6 chance for a single ball to drop into 1st urn.

That means there's a 5:6 chance of it getting in any of the others, which is the desired result.

Being 15 balls, that means the probability for all of them to land on different than 1st one is (5:6)^15 or 0.0649%.

Note that the not integer distribution is irrelevant because ball #2 has no interdependency or information on how ball#1 landed and so forth. Each drop is an independent action with a statistical chance result.

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