Conditional probability & r balls randomly distributed in n cells

[jfierro]

Conditional probability & "r balls randomly distributed in n cells"

Homework Statement

I'm posting this in hope that someone can give me a correct interpretation of the following problem (problem V.8 of Feller's Introduction to probability theory and its applications VOL I):

8. 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. Verify this numerically using table 1 of II,5.

Homework Equations

Conditional probability:
P\left \{A|B\right \} = \frac{P\left \{AB\right \}}{P\left \{B\right \}}
Number of ways of distributing r indistinguishable balls in n cells:
\binom{n+r-1}{r}
Number of ways of distributing r indistinguishable balls in n cells and no cell remaining empty:
\binom{r-1}{n-1}

The Attempt at a Solution

I cannot arrive at the 1/4 figure, so I thought this may be due to a misunderstanding of the problem statement. The way I interpreted the problem is saying that the new sample space is the same as if any 2 of the seven cells turn out to be empty, giving 315 possible arrangements:
\binom{7}{2}\binom{7-1}{5-1} = 315
With this scheme, only one cell of the remaining 5 can be triply occupied, because that leaves us with distributing the remaining 4 balls in the remaining 4 cells and no cell remaining empty. If we ignore the common factor 7 choose 2 = 21:
\frac{5}{\binom{7-1}{5-1}} = \frac{1}{3}

 

[Ray Vickson]

Homework Statement

I'm posting this in hope that someone can give me a correct interpretation of the following problem (problem V.8 of Feller's Introduction to probability theory and its applications VOL I):

8. 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. Verify this numerically using table 1 of II,5.

Homework Equations

Conditional probability:
P\left \{A|B\right \} = \frac{P\left \{AB\right \}}{P\left \{B\right \}}
Number of ways of distributing r indistinguishable balls in n cells:
\binom{n+r-1}{r}
Number of ways of distributing r indistinguishable balls in n cells and no cell remaining empty:
\binom{r-1}{n-1}

The Attempt at a Solution

I cannot arrive at the 1/4 figure, so I thought this may be due to a misunderstanding of the problem statement. The way I interpreted the problem is saying that the new sample space is the same as if any 2 of the seven cells turn out to be empty, giving 315 possible arrangements:
\binom{7}{2}\binom{7-1}{5-1} = 315
With this scheme, only one cell of the remaining 5 can be triply occupied, because that leaves us with distributing the remaining 4 balls in the remaining 4 cells and no cell remaining empty. If we ignore the common factor 7 choose 2 = 21:
\frac{5}{\binom{7-1}{5-1}} = \frac{1}{3}

Presumably you are supposed to come up with some clever direct argument, but barring that, Feller's hint will give you what you want (except for roundoff errors---which can be eliminated by going to exact rational expressions). Just consult Table I of II,5. What entries have exactly two cells empty? Among those, what entries have have triple entry cells?

RGV

 

[jfierro]

Thanks, now I got the correct answer. I guess my mistake was approaching the problem from "the number of indistinguishable configurations", ignoring that some of those configurations would appear more often than others.