# 7 balls are distributed randomly in 7 cells. If 2 cells are empty, show that the conditional probability of a triple occupancy equals 1/4

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?

• Are exactly two cells empty, or at least two? Is a triple occupancy exactly three balls in one cell, or at least three? Is "a triple occupancy" exactly one triple occupancy, or at least one? – Karl Mar 1 at 4:16
• There are exactly 2 cells empty and exactly one triple occupancy i think I will update with the exact statement from the problem – Sam Haze Mar 1 at 4:22

Let $$T$$ be the event that there is a triple occupancy, and let $$E$$ be the event that exactly two cells are empty. We are asked for $$\Pr(T|E)=\frac{\Pr(T\cap E)}{\Pr(E)}$$

The number of distributions with exactly two empty cells and a triple occupancy is $$7\binom73\binom624!=88,200\tag1$$ There are $$7$$ ways to choose the cell with three balls, $$\binom73$$ ways to choose the balls to go into it, $$\binom62$$ ways to choose the two empty cells, and $$4!$$ to distribute the remaining balls into the remaining cells.

To compute the number of distributions with exactly two empty cells, we use the principle of inclusion and exclusion to get $$\binom 72\left(5^7-\binom514^4+\binom523^7-\binom532^7+\binom541^7\right)=352,800\tag2$$

There are $$\binom72$$ choices for the two empty cells, and $$5^7$$ ways to distribute the balls into the other $$5$$ cells. Now, we must subtract the distributions that use only $$4$$ of the cells, that is $$\binom514^7$$. We have subtracted distributions that use only $$3$$ of the cells twice, so we must add them back in, and so on.

Dividing $$(1)$$ by $$(2)$$ does indeed give $$\frac14$$.

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

= $$\dfrac{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2}}{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2} + \dbinom{7}{2,2,1,1,1,0,0}\dbinom{7}{2,3,2}} = 1/4$$

• Using Stirling numbers of the second kind simplifies computations a lot, if you have recourse to the tables. If not, I find using the multinomial coefficient the most routine (hence less error prone) way of computing the results of dice throws or analogous problems. – true blue anil Mar 1 at 7:21

Using Stirling Number of the second kind -

If we take any $$5$$ cells, number of ways to distribute $$7$$ balls such that none of the cells are empty is given by $$StirlingS2[7,5] = 140$$, without distinguishing between cells.

Now we consider cases where one of the cells has $$3$$ balls, which is $$\displaystyle {7 \choose 3} = 35$$ ways of choosing $$3$$ balls and rest $$4$$ balls go one each in remaining $$4$$ cells.

So the desired probability $$\displaystyle = \frac{35}{140} = \frac{1}{4}$$.