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What is the probability that exactly one box remains empty
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Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $frac37$ (mcq)
Can you tell me how should I proceed then?
probability
edited Jul 14 at 23:24
Key Flex
4,461525
asked Jul 14 at 22:43
Ri Ray
51
add a comment |Â
up vote
0
down vote
favorite
Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $frac37$ (mcq)
Can you tell me how should I proceed then?
probability
edited Jul 14 at 23:24
Key Flex
4,461525
asked Jul 14 at 22:43
Ri Ray
51
choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $frac37$ (mcq)
Can you tell me how should I proceed then?
probability
edited Jul 14 at 23:24
Key Flex
4,461525
asked Jul 14 at 22:43
Ri Ray
51
Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $frac37$ (mcq)
Can you tell me how should I proceed then?
probability
edited Jul 14 at 23:24
Key Flex
4,461525
asked Jul 14 at 22:43
Ri Ray
51
edited Jul 14 at 23:24
Key Flex
4,461525
edited Jul 14 at 23:24
Key Flex
4,461525
edited Jul 14 at 23:24
Key Flex
4,461525
4,461525
asked Jul 14 at 22:43
Ri Ray
51
asked Jul 14 at 22:43
Ri Ray
51
asked Jul 14 at 22:43
Ri Ray
51
51
choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02
add a comment |Â
choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02
choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02
choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2dbinomn2$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $dbinomn+n-1n$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $dfracn(n-1)dbinom2n-1n$.
Can you now solve the problem?
answered Jul 14 at 23:01
Key Flex
4,461525
add a comment |Â
up vote
1
down vote
Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$beginalign
S_1 &= binom41 left( frac34 right)^5 \
S_2 &= binom42 left( frac24 right)^5 \
S_3 &= binom43 left( frac14 right)^5 \
S_4 &= 0
endalign$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_[m] = S_m - binomm+1m S_m+1 + binomm+2m S_m+2 - dots pm binomNm S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_[1] = S_1 - binom21 S_2 + binom31 S_3 - binom41 S_4= boxed0.585938$$
answered Jul 15 at 13:12
awkward
5,14111021
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
add a comment |Â
up vote
0
down vote
I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $frac5!2!$ ball arrangements $= frac4!3!cdot frac5!2! = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $frac5!3!$ and $frac5!2!cdot 2!$ ball arrangements $= (frac3!2!cdot frac5!3!+frac3!2!cdot frac5!2!cdot 2!)cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $frac5!4!$ and $frac5!3!cdot 2!$ ball arrangements $=(2!cdot frac5!4!+2!cdot frac5!3!cdot 2!)cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 textempty) = frac600240+600+180+4 = frac6001024 = .58594$
answered Jul 14 at 23:58
Phil H
1,8442311
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2dbinomn2$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $dbinomn+n-1n$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $dfracn(n-1)dbinom2n-1n$.
Can you now solve the problem?
answered Jul 14 at 23:01
Key Flex
4,461525
add a comment |Â
up vote
2
down vote
accepted
Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2dbinomn2$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $dbinomn+n-1n$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $dfracn(n-1)dbinom2n-1n$.
Can you now solve the problem?
answered Jul 14 at 23:01
Key Flex
4,461525
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2dbinomn2$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $dbinomn+n-1n$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $dfracn(n-1)dbinom2n-1n$.
Can you now solve the problem?
answered Jul 14 at 23:01
Key Flex
4,461525
Let $A$ be the empty box, and $B$ be the box containing $2$ balls and $C$ is $n-2$ boxes containing $1$ ball. So, there are $2dbinomn2$ ways of arranging the letter sequence $ABC.....C$. Then the size of the sample is $dbinomn+n-1n$ because it is equivalent to the number of ways of placing $n$ $0$'s and $n-1$ $1$'s in the order. So the probability is $dfracn(n-1)dbinom2n-1n$.
Can you now solve the problem?
answered Jul 14 at 23:01
Key Flex
4,461525
answered Jul 14 at 23:01
Key Flex
4,461525
answered Jul 14 at 23:01
Key Flex
4,461525
answered Jul 14 at 23:01
Key Flex
4,461525
4,461525
add a comment |Â
add a comment |Â
up vote
1
down vote
Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$beginalign
S_1 &= binom41 left( frac34 right)^5 \
S_2 &= binom42 left( frac24 right)^5 \
S_3 &= binom43 left( frac14 right)^5 \
S_4 &= 0
endalign$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_[m] = S_m - binomm+1m S_m+1 + binomm+2m S_m+2 - dots pm binomNm S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_[1] = S_1 - binom21 S_2 + binom31 S_3 - binom41 S_4= boxed0.585938$$
answered Jul 15 at 13:12
awkward
5,14111021
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
add a comment |Â
up vote
1
down vote
Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$beginalign
S_1 &= binom41 left( frac34 right)^5 \
S_2 &= binom42 left( frac24 right)^5 \
S_3 &= binom43 left( frac14 right)^5 \
S_4 &= 0
endalign$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_[m] = S_m - binomm+1m S_m+1 + binomm+2m S_m+2 - dots pm binomNm S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_[1] = S_1 - binom21 S_2 + binom31 S_3 - binom41 S_4= boxed0.585938$$
answered Jul 15 at 13:12
awkward
5,14111021
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$beginalign
S_1 &= binom41 left( frac34 right)^5 \
S_2 &= binom42 left( frac24 right)^5 \
S_3 &= binom43 left( frac14 right)^5 \
S_4 &= 0
endalign$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_[m] = S_m - binomm+1m S_m+1 + binomm+2m S_m+2 - dots pm binomNm S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_[1] = S_1 - binom21 S_2 + binom31 S_3 - binom41 S_4= boxed0.585938$$
answered Jul 15 at 13:12
awkward
5,14111021
Here is a solution by way of an extension of the Principle of Inclusion / Exclusion (PIE).
Say that an arrangement of balls in the boxes has "Property $i$" if box $i$ is empty, for $i=1,2,3,4$. Let $S_j$ be the total of the probabilities of all the arrangements with $j$ of the properties, for $j=1,2,3,4$. Then
$$beginalign
S_1 &= binom41 left( frac34 right)^5 \
S_2 &= binom42 left( frac24 right)^5 \
S_3 &= binom43 left( frac14 right)^5 \
S_4 &= 0
endalign$$
An extension of PIE states that the probability that exactly $m$ of $N$ events will occur is
$$P_[m] = S_m - binomm+1m S_m+1 + binomm+2m S_m+2 - dots pm binomNm S_N$$
Reference: An Introduction to Probability Theory and Its Applications, Volume I, Third Edition, by William Feller, Section IV.3.
Ours is the case $m=1$, $N=4$. So the probability that exactly one box is empty is
$$P_[1] = S_1 - binom21 S_2 + binom31 S_3 - binom41 S_4= boxed0.585938$$
answered Jul 15 at 13:12
awkward
5,14111021
answered Jul 15 at 13:12
awkward
5,14111021
answered Jul 15 at 13:12
awkward
5,14111021
answered Jul 15 at 13:12
awkward
5,14111021
5,14111021
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
add a comment |Â
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
There doesn't appear to be a consideration of zero empty boxes in your solution. I'm not familiar with this solution method so excuse me if this is an obviously incorrect observation.
– Phil H
Jul 15 at 16:59
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
@PhilH The "exactly one" part of the PIE extension takes care of the non-empty boxes. If exactly one box is empty, then the others are non-empty.
– awkward
Jul 16 at 0:25
add a comment |Â
up vote
0
down vote
I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $frac5!2!$ ball arrangements $= frac4!3!cdot frac5!2! = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $frac5!3!$ and $frac5!2!cdot 2!$ ball arrangements $= (frac3!2!cdot frac5!3!+frac3!2!cdot frac5!2!cdot 2!)cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $frac5!4!$ and $frac5!3!cdot 2!$ ball arrangements $=(2!cdot frac5!4!+2!cdot frac5!3!cdot 2!)cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 textempty) = frac600240+600+180+4 = frac6001024 = .58594$
answered Jul 14 at 23:58
Phil H
1,8442311
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
add a comment |Â
up vote
0
down vote
I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $frac5!2!$ ball arrangements $= frac4!3!cdot frac5!2! = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $frac5!3!$ and $frac5!2!cdot 2!$ ball arrangements $= (frac3!2!cdot frac5!3!+frac3!2!cdot frac5!2!cdot 2!)cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $frac5!4!$ and $frac5!3!cdot 2!$ ball arrangements $=(2!cdot frac5!4!+2!cdot frac5!3!cdot 2!)cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 textempty) = frac600240+600+180+4 = frac6001024 = .58594$
answered Jul 14 at 23:58
Phil H
1,8442311
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $frac5!2!$ ball arrangements $= frac4!3!cdot frac5!2! = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $frac5!3!$ and $frac5!2!cdot 2!$ ball arrangements $= (frac3!2!cdot frac5!3!+frac3!2!cdot frac5!2!cdot 2!)cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $frac5!4!$ and $frac5!3!cdot 2!$ ball arrangements $=(2!cdot frac5!4!+2!cdot frac5!3!cdot 2!)cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 textempty) = frac600240+600+180+4 = frac6001024 = .58594$
answered Jul 14 at 23:58
Phil H
1,8442311
I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $frac5!2!$ ball arrangements $= frac4!3!cdot frac5!2! = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $frac5!3!$ and $frac5!2!cdot 2!$ ball arrangements $= (frac3!2!cdot frac5!3!+frac3!2!cdot frac5!2!cdot 2!)cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $frac5!4!$ and $frac5!3!cdot 2!$ ball arrangements $=(2!cdot frac5!4!+2!cdot frac5!3!cdot 2!)cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 textempty) = frac600240+600+180+4 = frac6001024 = .58594$
answered Jul 14 at 23:58
Phil H
1,8442311
edited Jul 16 at 2:34
answered Jul 14 at 23:58
Phil H
1,8442311
answered Jul 14 at 23:58
Phil H
1,8442311
answered Jul 14 at 23:58
Phil H
1,8442311
1,8442311
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
add a comment |Â
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
You seem to have assumed that each of the arrangements listed is equally likely, which is false if each box is chosen independently and with equal probability.
– awkward
Jul 15 at 13:23
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
@awkward Yes, thinking about it a little more something isn't right. I can't see what it is though. Randomly placing each of 5 balls into 4 boxes seems to me to be a probability of $(frac14)^5$ for any single unique outcome. What am I missing?
– Phil H
Jul 15 at 16:56
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
First, there are $4^5$ ways to assign the 5 balls to 4 boxes, all of which we assume are equally likely. Consider the 0 boxes empty case: 1,1,1,2 can be arranged in 4 ways. But for each of these ways, the balls can be assigned to the boxes in $5! / 2! = 60$ ways. Think of listing the balls from 1 to 5, then writing the number of the relevant box under the number of the ball. For example, if we have one ball in each of boxes 1,2,3 and two balls in box 4, then there are $60$ ways to permute 1,2,3,4,4. So the probability of 0 boxes empty is $4 times 60 / 4^5$.
– awkward
Jul 16 at 1:42
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
Thanks awkward. I corrected my counting errors in my answer. I did read up on inclusion/exclusion probability which is based on overlapping quantities and hence adding and subtracting overlaps due to under or over counting. Thanks for your help.
– Phil H
Jul 16 at 2:40
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
You are welcome. Inclusion / exclusion is well worth studying, because its use greatly simplifies many problems in probability and combinatorics.
– awkward
Jul 16 at 12:12
add a comment |Â
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choosing the empty box in 4 ways is problematic because there will be cases where more than 1 box is empty. divide your cases as follows: how many arrangements are there if there are exactly 1, 2, or 3 empty boxes?
– Barycentric_Bash
Jul 14 at 23:02