2
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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? $\endgroup$ – Karl Mar 1 at 4:16
  • $\begingroup$ There are exactly 2 cells empty and exactly one triple occupancy i think I will update with the exact statement from the problem $\endgroup$ – Sam Haze Mar 1 at 4:22
2
$\begingroup$

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 $\frac14$.

$\endgroup$
3
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – true blue anil Mar 1 at 7:21
2
$\begingroup$

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 $\displaystyle = \frac{35}{140} = \frac{1}{4}$.

$\endgroup$

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.