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What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
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10 manuscripts are placed in 30 folders and each manuscript needs 3 folders. What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
Here is my solution
Let $mathcalM_i$ denote event of i-th manuscript being complete. Let us find $$mathbbP(bigcup_i=1^10mathcalM_i)$$
To do that me may find $mathbbP(mathcalM_1)$ which denotes of 1st manuscript being complete that is cotaining 3 folders of 1st manuscrpt and 3 of any other folder; therefore $$mathbbP(mathcalM_1) = fracbinom273binom306$$
and by inclusion exclusion we know $sum_i=1^10mathbbP(mathcalM_i) = binom101* mathbbP(mathcalM_1)$.
Following the same logic we know that 2 mansucripts can be complete if half of 6 folders belong to one manuscript and other half to another and there is only one such combination for each case;$$sum_i<jmathbbP(mathcalM_icapmathcalM_j) = binom102frac1binom306 = frac45binom306$$; we cannot have 3 or more complete manuscripts by selecting only 6 folders thus by inclusion exclusion;
$$mathbbP(bigcup_i=1^10mathcalM_i) = frac10binom273-45binom306 $$
and the needed result is $1 - mathbbP(bigcup_i=1^10mathcalM_i) = 0.950814702$
Anything wrong in my logic?
probability probability-theory proof-verification inclusion-exclusion
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
add a comment |Â
up vote
3
down vote
favorite
10 manuscripts are placed in 30 folders and each manuscript needs 3 folders. What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
Here is my solution
Let $mathcalM_i$ denote event of i-th manuscript being complete. Let us find $$mathbbP(bigcup_i=1^10mathcalM_i)$$
To do that me may find $mathbbP(mathcalM_1)$ which denotes of 1st manuscript being complete that is cotaining 3 folders of 1st manuscrpt and 3 of any other folder; therefore $$mathbbP(mathcalM_1) = fracbinom273binom306$$
and by inclusion exclusion we know $sum_i=1^10mathbbP(mathcalM_i) = binom101* mathbbP(mathcalM_1)$.
Following the same logic we know that 2 mansucripts can be complete if half of 6 folders belong to one manuscript and other half to another and there is only one such combination for each case;$$sum_i<jmathbbP(mathcalM_icapmathcalM_j) = binom102frac1binom306 = frac45binom306$$; we cannot have 3 or more complete manuscripts by selecting only 6 folders thus by inclusion exclusion;
$$mathbbP(bigcup_i=1^10mathcalM_i) = frac10binom273-45binom306 $$
and the needed result is $1 - mathbbP(bigcup_i=1^10mathcalM_i) = 0.950814702$
Anything wrong in my logic?
probability probability-theory proof-verification inclusion-exclusion
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
1
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
1
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
10 manuscripts are placed in 30 folders and each manuscript needs 3 folders. What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
Here is my solution
Let $mathcalM_i$ denote event of i-th manuscript being complete. Let us find $$mathbbP(bigcup_i=1^10mathcalM_i)$$
To do that me may find $mathbbP(mathcalM_1)$ which denotes of 1st manuscript being complete that is cotaining 3 folders of 1st manuscrpt and 3 of any other folder; therefore $$mathbbP(mathcalM_1) = fracbinom273binom306$$
and by inclusion exclusion we know $sum_i=1^10mathbbP(mathcalM_i) = binom101* mathbbP(mathcalM_1)$.
Following the same logic we know that 2 mansucripts can be complete if half of 6 folders belong to one manuscript and other half to another and there is only one such combination for each case;$$sum_i<jmathbbP(mathcalM_icapmathcalM_j) = binom102frac1binom306 = frac45binom306$$; we cannot have 3 or more complete manuscripts by selecting only 6 folders thus by inclusion exclusion;
$$mathbbP(bigcup_i=1^10mathcalM_i) = frac10binom273-45binom306 $$
and the needed result is $1 - mathbbP(bigcup_i=1^10mathcalM_i) = 0.950814702$
Anything wrong in my logic?
probability probability-theory proof-verification inclusion-exclusion
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
10 manuscripts are placed in 30 folders and each manuscript needs 3 folders. What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
Here is my solution
Let $mathcalM_i$ denote event of i-th manuscript being complete. Let us find $$mathbbP(bigcup_i=1^10mathcalM_i)$$
To do that me may find $mathbbP(mathcalM_1)$ which denotes of 1st manuscript being complete that is cotaining 3 folders of 1st manuscrpt and 3 of any other folder; therefore $$mathbbP(mathcalM_1) = fracbinom273binom306$$
and by inclusion exclusion we know $sum_i=1^10mathbbP(mathcalM_i) = binom101* mathbbP(mathcalM_1)$.
Following the same logic we know that 2 mansucripts can be complete if half of 6 folders belong to one manuscript and other half to another and there is only one such combination for each case;$$sum_i<jmathbbP(mathcalM_icapmathcalM_j) = binom102frac1binom306 = frac45binom306$$; we cannot have 3 or more complete manuscripts by selecting only 6 folders thus by inclusion exclusion;
$$mathbbP(bigcup_i=1^10mathcalM_i) = frac10binom273-45binom306 $$
and the needed result is $1 - mathbbP(bigcup_i=1^10mathcalM_i) = 0.950814702$
Anything wrong in my logic?
probability probability-theory proof-verification inclusion-exclusion
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
edited Jul 15 at 22:57
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
asked Jul 15 at 22:29
Erik Hambardzumyan
1548
1548
I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
1
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
1
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02
add a comment |Â
I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
1
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
1
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02
I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
1
1
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
1
1
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02
add a comment |Â
1 Answer
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Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,ldots,10$) be defined similarly.
Then $S = cup_i=1^10 S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle,
beginalign
|S| &= sum_i=1^10 | S_i | - sum_i < j | S_i cap S_j| \
&= sum_i=1^10 binom273 - sum_i < j 1 \
&= 10 binom273 - 45.
endalign
The total number of sets of 6 folders is $binom306$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is
$$
p = frac10 binom273 - 45binom306.
$$
The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.
answered Jul 15 at 23:22
littleO
26.2k540102
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,ldots,10$) be defined similarly.
Then $S = cup_i=1^10 S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle,
beginalign
|S| &= sum_i=1^10 | S_i | - sum_i < j | S_i cap S_j| \
&= sum_i=1^10 binom273 - sum_i < j 1 \
&= 10 binom273 - 45.
endalign
The total number of sets of 6 folders is $binom306$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is
$$
p = frac10 binom273 - 45binom306.
$$
The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.
answered Jul 15 at 23:22
littleO
26.2k540102
add a comment |Â
up vote
2
down vote
accepted
Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,ldots,10$) be defined similarly.
Then $S = cup_i=1^10 S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle,
beginalign
|S| &= sum_i=1^10 | S_i | - sum_i < j | S_i cap S_j| \
&= sum_i=1^10 binom273 - sum_i < j 1 \
&= 10 binom273 - 45.
endalign
The total number of sets of 6 folders is $binom306$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is
$$
p = frac10 binom273 - 45binom306.
$$
The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.
answered Jul 15 at 23:22
littleO
26.2k540102
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,ldots,10$) be defined similarly.
Then $S = cup_i=1^10 S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle,
beginalign
|S| &= sum_i=1^10 | S_i | - sum_i < j | S_i cap S_j| \
&= sum_i=1^10 binom273 - sum_i < j 1 \
&= 10 binom273 - 45.
endalign
The total number of sets of 6 folders is $binom306$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is
$$
p = frac10 binom273 - 45binom306.
$$
The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.
answered Jul 15 at 23:22
littleO
26.2k540102
Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,ldots,10$) be defined similarly.
Then $S = cup_i=1^10 S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle,
beginalign
|S| &= sum_i=1^10 | S_i | - sum_i < j | S_i cap S_j| \
&= sum_i=1^10 binom273 - sum_i < j 1 \
&= 10 binom273 - 45.
endalign
The total number of sets of 6 folders is $binom306$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is
$$
p = frac10 binom273 - 45binom306.
$$
The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.
answered Jul 15 at 23:22
littleO
26.2k540102
answered Jul 15 at 23:22
littleO
26.2k540102
answered Jul 15 at 23:22
littleO
26.2k540102
answered Jul 15 at 23:22
littleO
26.2k540102
26.2k540102
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I don't think the question is clear enough to be answered. What does "each manuscript needs 3 folders" mean? Does it mean each manuscript is in 3 parts, and each folder can hold only one part? And are parts kept together as units without further splitting, so each folder has at most one manuscript part? Is it possible for a single folder to have pieces of two different manuscripts?
– Michael
Jul 15 at 22:42
@Michael My guess was that each manuscript is divided into 3 pieces each placed into separate folder
– Erik Hambardzumyan
Jul 15 at 22:46
1
I get the same answer as you by a different method, which suggests we are both correct.
– Michael
Jul 15 at 22:54
@Michael Glad to hear that, what is your method?
– Erik Hambardzumyan
Jul 15 at 22:55
1
Our denominators are the same, our numerators are also the same but my way is not as elegant: We could have $[1;1;1;1;1;1], [1;1;1;1;2],[1;1;2;2],[2;2;2]$ and so $$numerator=564570 = binom1063^6 + binom105(5)(3^5) + binom104binom423^4 + binom1033^3$$
– Michael
Jul 15 at 23:02