- #1
jfierro
- 20
- 1
Conditional probability & "r balls randomly distributed in n cells"
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.
Conditional probability:
[tex]P\left \{A|B\right \} = \frac{P\left \{AB\right \}}{P\left \{B\right \}}[/tex]
Number of ways of distributing r indistinguishable balls in n cells:
[tex]\binom{n+r-1}{r}[/tex]
Number of ways of distributing r indistinguishable balls in n cells and no cell remaining empty:
[tex]\binom{r-1}{n-1}[/tex]
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:
[tex]\binom{7}{2}\binom{7-1}{5-1} = 315[/tex]
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:
[tex]\frac{5}{\binom{7-1}{5-1}} = \frac{1}{3}[/tex]
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:
[tex]P\left \{A|B\right \} = \frac{P\left \{AB\right \}}{P\left \{B\right \}}[/tex]
Number of ways of distributing r indistinguishable balls in n cells:
[tex]\binom{n+r-1}{r}[/tex]
Number of ways of distributing r indistinguishable balls in n cells and no cell remaining empty:
[tex]\binom{r-1}{n-1}[/tex]
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:
[tex]\binom{7}{2}\binom{7-1}{5-1} = 315[/tex]
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:
[tex]\frac{5}{\binom{7-1}{5-1}} = \frac{1}{3}[/tex]