So I have
- $10$ balls, and $6$ urns
All of the balls and urns are numbered/labeled. This tells me that everything is unique.
The question is, what is the probability that none of the urns are empty.
Attempt
$Pr($None are empty$)= Pr($all are full$)=Pr($I not empty$)*Pr($II not empty$)*...Pr($VI not empty) In a previous problem I found the probability of Urn $I$ to be empty as $\frac{5^{10}}{6^{10}}$ So the probability of Urn $I$ not empty is 1-$\frac{5^{10}}{6^{10}}$. Since the probability of any one urn to be empty is the same. its 6*(1-$\frac{5^{10}}{6^{10}}$). This doesn't feel right based on the answer solutions in this question appears to be similar however I am not sure how to extend that here.
I am not sure how to extend the answer solution there to this case either. Ball 1 has only 1 choices, urn $I$. Ball 2: has two choices($I, II) =>$ ($0.5$), ... Ball 6: =>$\frac{1}{6}$. However ball 7, 8, 9, 10 now has 6 choices. So does this mean that the $Pr($none are empty)=($\frac{1}{6!*6^3}$)
By inclusion exclusion principle the probability in question is: $$ \frac1{6^{10}}\sum_{i=0}^6(-1)^i\binom{6}i (6-i)^{10}\equiv\frac{6!}{6^{10}}{10\brace 6}\approx0.27, $$ where ${m \brace n} $ is the Stirling number of the second kind.