This problem comes from Feller's Introduction to probability. and it goes like:
"Seven balls are distributed randomly in seven cells. Given that two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals 1/4"
So far i have found out that all of the distributions that form a triple are given by: ${7 \choose 3}(4!) = \frac{7!}{3!}$.
Yet, i am not sure of how to go on, because there are $7^7$ total distributions, but clearly if you divide these two quantities the result is not even close to $1/4$.
I have also tried to go on by considering that the only occupancy numbers in the distribution can be $0,1,3$ so the total number of distributions that coincide with the ocuppancy numbers $3,1,1,1,1,0,0$ in some order are given by: $\frac{7!}{3!}\frac{7!}{3!4!}$. But if you divide this by $7^7$ the result is still no close to $1/4$.
And also when you consider the fact of conditional probability the results only get worse. So what can i do?
Best Answer
To me, it looks like throwing a $7$ sided die $7$ times, with the the desired conditional probability being
n(1 cell has triple occupancy with 2 cells unoccupied) $\div$ n(2 cells unoccupied)
Using the multinomial coefficient, we get
= $\dfrac{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2}}{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2} + \dbinom{7}{2,2,1,1,1,0,0}\dbinom{7}{2,3,2}} = 1/4$