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Calculating the probability of an element being in 2 subsets [closed]
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I've just started to learn about "Probability-Theory" and came across this problem that i want to calculate.
I've given a total amount of 28 million elements. Following information is given: In total there are 2562 elements with Attribute A and 3 elements with the Attribute B. Now how can i calculate the probability that one randomly drawn element has BOTH Attributes A and B
And how does the probability behave if I draw X elements (without replacement).
I've also drawn a little picture to visualize my problem.
Thanks in advance,
Koapsii
Edit: The Attributes are randomly distributed. So:
Step 1: Go through the 28 mil and distribute Attribute A randomly
Step 2: Go through the 28 mil and distribute Attribute B randomly
Step 3: Calculate amount/likelihood that an element has Attribute A AND B
The problem is about IT Security. In total there are 28 mil software entities, researches estimated that there are ~2562 hidden vulnerabilities (so called ZeroDay vulnerabilities) in these software entities and now I want to find out what's the chance that one of my 3 software entities is affected by those 2562 hidden vulnerabilities. There is no way i can look that up, because this is exactly what i want to know. The chance of being affected
probability probability-theory education
asked Jul 25 at 14:55
koapsi
206
closed as off-topic by amWhy, José Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel Jul 26 at 1:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Jos̩ Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel
 |Â
show 2 more comments
up vote
-1
down vote
favorite
I've just started to learn about "Probability-Theory" and came across this problem that i want to calculate.
I've given a total amount of 28 million elements. Following information is given: In total there are 2562 elements with Attribute A and 3 elements with the Attribute B. Now how can i calculate the probability that one randomly drawn element has BOTH Attributes A and B
And how does the probability behave if I draw X elements (without replacement).
I've also drawn a little picture to visualize my problem.
Thanks in advance,
Koapsii
Edit: The Attributes are randomly distributed. So:
Step 1: Go through the 28 mil and distribute Attribute A randomly
Step 2: Go through the 28 mil and distribute Attribute B randomly
Step 3: Calculate amount/likelihood that an element has Attribute A AND B
The problem is about IT Security. In total there are 28 mil software entities, researches estimated that there are ~2562 hidden vulnerabilities (so called ZeroDay vulnerabilities) in these software entities and now I want to find out what's the chance that one of my 3 software entities is affected by those 2562 hidden vulnerabilities. There is no way i can look that up, because this is exactly what i want to know. The chance of being affected
probability probability-theory education
asked Jul 25 at 14:55
koapsi
206
closed as off-topic by amWhy, José Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel Jul 26 at 1:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Jos̩ Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05
 |Â
show 2 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I've just started to learn about "Probability-Theory" and came across this problem that i want to calculate.
I've given a total amount of 28 million elements. Following information is given: In total there are 2562 elements with Attribute A and 3 elements with the Attribute B. Now how can i calculate the probability that one randomly drawn element has BOTH Attributes A and B
And how does the probability behave if I draw X elements (without replacement).
I've also drawn a little picture to visualize my problem.
Thanks in advance,
Koapsii
Edit: The Attributes are randomly distributed. So:
Step 1: Go through the 28 mil and distribute Attribute A randomly
Step 2: Go through the 28 mil and distribute Attribute B randomly
Step 3: Calculate amount/likelihood that an element has Attribute A AND B
The problem is about IT Security. In total there are 28 mil software entities, researches estimated that there are ~2562 hidden vulnerabilities (so called ZeroDay vulnerabilities) in these software entities and now I want to find out what's the chance that one of my 3 software entities is affected by those 2562 hidden vulnerabilities. There is no way i can look that up, because this is exactly what i want to know. The chance of being affected
probability probability-theory education
asked Jul 25 at 14:55
koapsi
206
I've just started to learn about "Probability-Theory" and came across this problem that i want to calculate.
I've given a total amount of 28 million elements. Following information is given: In total there are 2562 elements with Attribute A and 3 elements with the Attribute B. Now how can i calculate the probability that one randomly drawn element has BOTH Attributes A and B
And how does the probability behave if I draw X elements (without replacement).
I've also drawn a little picture to visualize my problem.
Thanks in advance,
Koapsii
Edit: The Attributes are randomly distributed. So:
Step 1: Go through the 28 mil and distribute Attribute A randomly
Step 2: Go through the 28 mil and distribute Attribute B randomly
Step 3: Calculate amount/likelihood that an element has Attribute A AND B
The problem is about IT Security. In total there are 28 mil software entities, researches estimated that there are ~2562 hidden vulnerabilities (so called ZeroDay vulnerabilities) in these software entities and now I want to find out what's the chance that one of my 3 software entities is affected by those 2562 hidden vulnerabilities. There is no way i can look that up, because this is exactly what i want to know. The chance of being affected
probability probability-theory education
asked Jul 25 at 14:55
koapsi
206
edited Jul 25 at 15:37
asked Jul 25 at 14:55
koapsi
206
asked Jul 25 at 14:55
koapsi
206
asked Jul 25 at 14:55
koapsi
206
206
closed as off-topic by amWhy, José Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel Jul 26 at 1:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Jos̩ Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel
closed as off-topic by amWhy, José Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel Jul 26 at 1:02
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РamWhy, Jos̩ Carlos Santos, Leucippus, Adrian Keister, Parcly Taxel
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05
 |Â
show 2 more comments
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your calculation assumes that having attribute A and attribute B are independent. It would apply if you went through the elements and gave them attribute A with probability $frac 256228,000,000$ and attribute B with probability $frac 328,000,000$. Given that you have the sample and exact numbers with each attribute they can't be independent. If there is one that has both attributes, given a $B$ the chance it is an $A$ is (at least) $frac 13$. If there are none with both attributes the chance that a $B$ is also an $A$ is $0$. Both of these are different from $frac 256228,000,000$.
answered Jul 25 at 15:07
Ross Millikan
275k21186351
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your calculation assumes that having attribute A and attribute B are independent. It would apply if you went through the elements and gave them attribute A with probability $frac 256228,000,000$ and attribute B with probability $frac 328,000,000$. Given that you have the sample and exact numbers with each attribute they can't be independent. If there is one that has both attributes, given a $B$ the chance it is an $A$ is (at least) $frac 13$. If there are none with both attributes the chance that a $B$ is also an $A$ is $0$. Both of these are different from $frac 256228,000,000$.
answered Jul 25 at 15:07
Ross Millikan
275k21186351
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
 |Â
show 1 more comment
up vote
1
down vote
accepted
Your calculation assumes that having attribute A and attribute B are independent. It would apply if you went through the elements and gave them attribute A with probability $frac 256228,000,000$ and attribute B with probability $frac 328,000,000$. Given that you have the sample and exact numbers with each attribute they can't be independent. If there is one that has both attributes, given a $B$ the chance it is an $A$ is (at least) $frac 13$. If there are none with both attributes the chance that a $B$ is also an $A$ is $0$. Both of these are different from $frac 256228,000,000$.
answered Jul 25 at 15:07
Ross Millikan
275k21186351
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your calculation assumes that having attribute A and attribute B are independent. It would apply if you went through the elements and gave them attribute A with probability $frac 256228,000,000$ and attribute B with probability $frac 328,000,000$. Given that you have the sample and exact numbers with each attribute they can't be independent. If there is one that has both attributes, given a $B$ the chance it is an $A$ is (at least) $frac 13$. If there are none with both attributes the chance that a $B$ is also an $A$ is $0$. Both of these are different from $frac 256228,000,000$.
answered Jul 25 at 15:07
Ross Millikan
275k21186351
Your calculation assumes that having attribute A and attribute B are independent. It would apply if you went through the elements and gave them attribute A with probability $frac 256228,000,000$ and attribute B with probability $frac 328,000,000$. Given that you have the sample and exact numbers with each attribute they can't be independent. If there is one that has both attributes, given a $B$ the chance it is an $A$ is (at least) $frac 13$. If there are none with both attributes the chance that a $B$ is also an $A$ is $0$. Both of these are different from $frac 256228,000,000$.
answered Jul 25 at 15:07
Ross Millikan
275k21186351
answered Jul 25 at 15:07
Ross Millikan
275k21186351
answered Jul 25 at 15:07
Ross Millikan
275k21186351
answered Jul 25 at 15:07
Ross Millikan
275k21186351
275k21186351
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
 |Â
show 1 more comment
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
Well, i only know that there are 2562 elements with Attribute A and 3 elements with Attribute B within these 28 million elements. So i assume that they are independent. It could be that all 3 elements with Attribute B also have Attribute A or that none of the elements with Attribute B have Attribute A. But i somehow need the amount of Attributes that have A and B. Or atleast how likely it is that there is an element with Attribute A and B
– koapsi
Jul 25 at 15:17
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
If you randomly pick without replacement $2562$ to give $A$ to and then pick without replacement $3$ to give $B$ to, the chance there is a match is very close to $frac 3cdot 256228,000,000$. To be exact, go through the $B$s, compute the chance that none is an A, and subtract from $1$. It will be higher because you use up some non-As, so $1-(frac 28,000,000-256228,000,000cdotfrac 27,999,999-256227,999,999cdotfrac 27,999,998-256227,999,998)$.
– Ross Millikan
Jul 25 at 15:24
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
Yes, i randomly pick 2562 elements without replacement and give them Attribute A and then I randomly pick 3 elements without replacement and give them Attribute B. After the Attribute Distribution i take exactly one element out of the 28 million and therefor i want to know the chance that this certain element has both Attributes. As far as I understand your formula, it calculates the chance of 1 - Picking an element without Attribute. But does this consider that i only pick ONE element after the attribute distribution?
– koapsi
Jul 25 at 16:12
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
No, this is the chance there is an element that is both an A and a B. It includes the chance there are two, but that is tiny. Then there is $1/28,000,000$ chance you pick that element
– Ross Millikan
Jul 25 at 16:23
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
Ohhhhh, now I understand. I guess you saved my week.
– koapsi
Jul 25 at 16:24
 |Â
show 1 more comment
You should clarify how many elements there are in a sample.
– gimusi
Jul 25 at 14:57
Well, as I've said: "one randomly drawn element", so the sample has 1 element to understand the idea behind the formula. And after that, I would like to know how the probability calculation behaves with X elements in the sample
– koapsi
Jul 25 at 15:00
You just asked this question, didn't you?
– saulspatz
Jul 25 at 15:00
@saulspatz yes I did, but people asked me to reopen the question, since I had some wrong formulations in it.
– koapsi
Jul 25 at 15:01
When that happens, it's better to edit the original question. This way, all the previous discussion is lost. Still, you have the same problems in this formulation: 1) we don't know how many objects have both attributes . 2) Your answer can't possibly be right. Look at the denominator. It's far greater than the number of objects in the sample space. There simply isn't enough information given to answer the question.
– saulspatz
Jul 25 at 15:05