Are two events with non zero probability of occurence and empty intersection independent [duplicate]
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This question already has an answer here:
The definition of independence is not intuitive
2 answers
Given two events $A, B$ of which the first is the outcome of flipping a coin and the second is the outcome of a weather forcasting, one will intuitively say that the two events are independent, since the outcome of $A$ in no way would effect the outcome of $B.$ So one could write $P(A|B)=P(A).$ But then I have a problem which is: from the independence must follow $P(Acap B)=P(A)P(B).$ I guess $Acap B=emptyset $ since I dont see what the two events can have in common and so $0=P(A)P(B),$ where $P(A), P(B)$ must not be zero.
What is wrong in these thoughts ? Are two events which have empty intersection independent if the probability of each is non zero ?
probability
asked Dec 31 '18 at 23:27
user249018user249018
435137
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marked as duplicate by Jyrki Lahtonen, Cesareo, KReiser, drhab probability
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Jan 1 at 17:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
The definition of independence is not intuitive
2 answers
Given two events $A, B$ of which the first is the outcome of flipping a coin and the second is the outcome of a weather forcasting, one will intuitively say that the two events are independent, since the outcome of $A$ in no way would effect the outcome of $B.$ So one could write $P(A|B)=P(A).$ But then I have a problem which is: from the independence must follow $P(Acap B)=P(A)P(B).$ I guess $Acap B=emptyset $ since I dont see what the two events can have in common and so $0=P(A)P(B),$ where $P(A), P(B)$ must not be zero.
What is wrong in these thoughts ? Are two events which have empty intersection independent if the probability of each is non zero ?
probability
asked Dec 31 '18 at 23:27
user249018user249018
435137
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marked as duplicate by Jyrki Lahtonen, Cesareo, KReiser, drhab probability
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Jan 1 at 17:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, I'm not sure I understand the confusion here.
$endgroup$
– Don Thousand
Dec 31 '18 at 23:30
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@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
$endgroup$
– user249018
Jan 1 at 0:00
1
$begingroup$
Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:07
$begingroup$
To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:14
1
$begingroup$
Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:19
add a comment |
$begingroup$
This question already has an answer here:
The definition of independence is not intuitive
2 answers
Given two events $A, B$ of which the first is the outcome of flipping a coin and the second is the outcome of a weather forcasting, one will intuitively say that the two events are independent, since the outcome of $A$ in no way would effect the outcome of $B.$ So one could write $P(A|B)=P(A).$ But then I have a problem which is: from the independence must follow $P(Acap B)=P(A)P(B).$ I guess $Acap B=emptyset $ since I dont see what the two events can have in common and so $0=P(A)P(B),$ where $P(A), P(B)$ must not be zero.
What is wrong in these thoughts ? Are two events which have empty intersection independent if the probability of each is non zero ?
probability
asked Dec 31 '18 at 23:27
user249018user249018
435137
$endgroup$
This question already has an answer here:
The definition of independence is not intuitive
2 answers
Given two events $A, B$ of which the first is the outcome of flipping a coin and the second is the outcome of a weather forcasting, one will intuitively say that the two events are independent, since the outcome of $A$ in no way would effect the outcome of $B.$ So one could write $P(A|B)=P(A).$ But then I have a problem which is: from the independence must follow $P(Acap B)=P(A)P(B).$ I guess $Acap B=emptyset $ since I dont see what the two events can have in common and so $0=P(A)P(B),$ where $P(A), P(B)$ must not be zero.
What is wrong in these thoughts ? Are two events which have empty intersection independent if the probability of each is non zero ?
This question already has an answer here:
The definition of independence is not intuitive
2 answers
probability
probability
asked Dec 31 '18 at 23:27
user249018user249018
435137
asked Dec 31 '18 at 23:27
user249018user249018
435137
asked Dec 31 '18 at 23:27
user249018user249018
435137
asked Dec 31 '18 at 23:27
user249018user249018
435137
asked Dec 31 '18 at 23:27
user249018user249018
435137
435137
marked as duplicate by Jyrki Lahtonen, Cesareo, KReiser, drhab probability
Users with the probability badge can single-handedly close probability questions as duplicates and reopen them as needed.
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Jan 1 at 17:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, Cesareo, KReiser, drhab probability
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Jan 1 at 17:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Yes, I'm not sure I understand the confusion here.
$endgroup$
– Don Thousand
Dec 31 '18 at 23:30
$begingroup$
@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
$endgroup$
– user249018
Jan 1 at 0:00
1
$begingroup$
Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:07
$begingroup$
To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:14
1
$begingroup$
Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:19
add a comment |
$begingroup$
Yes, I'm not sure I understand the confusion here.
$endgroup$
– Don Thousand
Dec 31 '18 at 23:30
$begingroup$
@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
$endgroup$
– user249018
Jan 1 at 0:00
1
$begingroup$
Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:07
$begingroup$
To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:14
1
$begingroup$
Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:19
$begingroup$
Yes, I'm not sure I understand the confusion here.
$endgroup$
– Don Thousand
Dec 31 '18 at 23:30
$begingroup$
Yes, I'm not sure I understand the confusion here.
$endgroup$
– Don Thousand
Dec 31 '18 at 23:30
$begingroup$
@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
$endgroup$
– user249018
Jan 1 at 0:00
$begingroup$
@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
$endgroup$
– user249018
Jan 1 at 0:00
1
1
$begingroup$
Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:07
$begingroup$
Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:07
$begingroup$
To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:14
$begingroup$
To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:14
1
1
$begingroup$
Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:19
$begingroup$
Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
$endgroup$
– Jyrki Lahtonen
Jan 1 at 5:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You are assuming that mutually exclusive events are necessarily independent. That is not the case as you have mentioned one is about $P(Acap B)=0$ and the other one about $P(Acap B) = P(A)P(B)$
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
$endgroup$
– user249018
Jan 1 at 0:41
1
$begingroup$
They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 1:19
$begingroup$
Main mistake is the assumption that the events are mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:40
add a comment |
$begingroup$
Suppose we have $P(A), P(B) > 0$ and $A cap B = emptyset implies P(A cap B) = 0$.
This is not what was call "independent," but rather "mutually exclusive" - only one of $A,B$ can occur. You can heuristically look at this is an extreme sort of dependence - they are dependent in that if one occurs, then the other cannot.
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
$endgroup$
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
$endgroup$
– user249018
Jan 1 at 0:07
$begingroup$
Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
$endgroup$
– Eevee Trainer
Jan 1 at 0:09
$begingroup$
Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
$endgroup$
– user249018
Jan 1 at 0:17
$begingroup$
It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:38
add a comment |
$begingroup$
No, such events are dependent (in extrem way). Events $A$ and it negation $A'$ are disjunct but they are (very) dependent. If one hapens the other does not.
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
$begingroup$
Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
$endgroup$
– user249018
Jan 1 at 0:13
add a comment |
$begingroup$
In a suitable model of reality in the form of a probability space $(Omega,mathcal A,P)$ the described events $A$ and $B$ will be both subsets of the $sigma$-algebra that is in use.
Further the probability measure that is in use will - if it indeed models reality on a suitable way - give: $$P(Acap B)=P(A)P(B)$$
The event $Acap B$ will - if $P(A)$ and $P(B)$ are both positive - not be empty.
If e.g. $(Omega_1,mathcal A_1,P_1)$ models the coin-flipping and $(Omega_2,mathcal A_2,P_2)$ models the weather forcasting then go for the product of both probability spaces.
answered Jan 1 at 11:36
drhabdrhab
103k545136
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add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are assuming that mutually exclusive events are necessarily independent. That is not the case as you have mentioned one is about $P(Acap B)=0$ and the other one about $P(Acap B) = P(A)P(B)$
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
$endgroup$
– user249018
Jan 1 at 0:41
1
$begingroup$
They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 1:19
$begingroup$
Main mistake is the assumption that the events are mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:40
add a comment |
$begingroup$
You are assuming that mutually exclusive events are necessarily independent. That is not the case as you have mentioned one is about $P(Acap B)=0$ and the other one about $P(Acap B) = P(A)P(B)$
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
$endgroup$
– user249018
Jan 1 at 0:41
1
$begingroup$
They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 1:19
$begingroup$
Main mistake is the assumption that the events are mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:40
add a comment |
$begingroup$
You are assuming that mutually exclusive events are necessarily independent. That is not the case as you have mentioned one is about $P(Acap B)=0$ and the other one about $P(Acap B) = P(A)P(B)$
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
You are assuming that mutually exclusive events are necessarily independent. That is not the case as you have mentioned one is about $P(Acap B)=0$ and the other one about $P(Acap B) = P(A)P(B)$
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 31 '18 at 23:36
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
$endgroup$
– user249018
Jan 1 at 0:41
1
$begingroup$
They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 1:19
$begingroup$
Main mistake is the assumption that the events are mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:40
add a comment |
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
$endgroup$
– user249018
Jan 1 at 0:41
1
$begingroup$
They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 1:19
$begingroup$
Main mistake is the assumption that the events are mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:40
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
$begingroup$
Many thanks. To put it short, are the events $A, B$ exclusive or indepenedent without taking into consideration my assumption that $Acap B=emptyset$ ?
$endgroup$
– user249018
Jan 1 at 0:09
1
1
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
$begingroup$
$P(Acap B)=0 $ is the mutual exclusive case in which the events are dependent unless both have zero probability.
$endgroup$
– Mohammad Riazi-Kermani
Jan 1 at 0:35
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Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
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– user249018
Jan 1 at 0:41
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Ok. But how can the two events $A, B$ (coin flipping and weather forcasting) be dependent ? This is the core of the problem which I dont understand.
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– user249018
Jan 1 at 0:41
1
1
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They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
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– Mohammad Riazi-Kermani
Jan 1 at 1:19
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They are independent and they are not mutually exclusive. If probability of rain in one day is 1/3 and you flip a coin then probability of both raining and getting a head is (1/3)(1/2)=1/6.
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– Mohammad Riazi-Kermani
Jan 1 at 1:19
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Main mistake is the assumption that the events are mutually exclusive.
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– drhab
Jan 1 at 11:40
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Main mistake is the assumption that the events are mutually exclusive.
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– drhab
Jan 1 at 11:40
add a comment |
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Suppose we have $P(A), P(B) > 0$ and $A cap B = emptyset implies P(A cap B) = 0$.
This is not what was call "independent," but rather "mutually exclusive" - only one of $A,B$ can occur. You can heuristically look at this is an extreme sort of dependence - they are dependent in that if one occurs, then the other cannot.
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
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I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
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– user249018
Jan 1 at 0:07
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Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
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– Eevee Trainer
Jan 1 at 0:09
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Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
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– user249018
Jan 1 at 0:17
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It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
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– drhab
Jan 1 at 11:38
add a comment |
$begingroup$
Suppose we have $P(A), P(B) > 0$ and $A cap B = emptyset implies P(A cap B) = 0$.
This is not what was call "independent," but rather "mutually exclusive" - only one of $A,B$ can occur. You can heuristically look at this is an extreme sort of dependence - they are dependent in that if one occurs, then the other cannot.
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
$endgroup$
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
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– user249018
Jan 1 at 0:07
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Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
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– Eevee Trainer
Jan 1 at 0:09
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Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
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– user249018
Jan 1 at 0:17
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It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:38
add a comment |
$begingroup$
Suppose we have $P(A), P(B) > 0$ and $A cap B = emptyset implies P(A cap B) = 0$.
This is not what was call "independent," but rather "mutually exclusive" - only one of $A,B$ can occur. You can heuristically look at this is an extreme sort of dependence - they are dependent in that if one occurs, then the other cannot.
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
$endgroup$
Suppose we have $P(A), P(B) > 0$ and $A cap B = emptyset implies P(A cap B) = 0$.
This is not what was call "independent," but rather "mutually exclusive" - only one of $A,B$ can occur. You can heuristically look at this is an extreme sort of dependence - they are dependent in that if one occurs, then the other cannot.
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
answered Dec 31 '18 at 23:31
Eevee TrainerEevee Trainer
8,51231439
8,51231439
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
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– user249018
Jan 1 at 0:07
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Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
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– Eevee Trainer
Jan 1 at 0:09
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Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
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– user249018
Jan 1 at 0:17
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It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
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– drhab
Jan 1 at 11:38
add a comment |
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
$endgroup$
– user249018
Jan 1 at 0:07
$begingroup$
Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
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– Eevee Trainer
Jan 1 at 0:09
$begingroup$
Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
$endgroup$
– user249018
Jan 1 at 0:17
$begingroup$
It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:38
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
$endgroup$
– user249018
Jan 1 at 0:07
$begingroup$
I wrote, $P(A|B)=P(A)$ and thus I concluded $A,B$ are independent. So no matter what my comment on $A, B$ was, are $A, B$ as events independent or not ? Is my Statement $Acap B= emptyset$ true or false ?
$endgroup$
– user249018
Jan 1 at 0:07
$begingroup$
Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
$endgroup$
– Eevee Trainer
Jan 1 at 0:09
$begingroup$
Recall: $$P(A|B) = frac{P(A cap B)}{P(B)}$$ If $P(A cap B) = 0$, $P(A|B) = 0$. However, $P(A) > 0$. Thus, $P(A|B) neq P(A)$.
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– Eevee Trainer
Jan 1 at 0:09
$begingroup$
Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
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– user249018
Jan 1 at 0:17
$begingroup$
Given the two events $A,B$ as described above is then $Acap B=emptyset $ false ?
$endgroup$
– user249018
Jan 1 at 0:17
$begingroup$
It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
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– drhab
Jan 1 at 11:38
$begingroup$
It can happen that the coin gives "heads" and the weather forcast gives "rain". So events like that are not mutually exclusive.
$endgroup$
– drhab
Jan 1 at 11:38
add a comment |
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No, such events are dependent (in extrem way). Events $A$ and it negation $A'$ are disjunct but they are (very) dependent. If one hapens the other does not.
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
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Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
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– user249018
Jan 1 at 0:13
add a comment |
$begingroup$
No, such events are dependent (in extrem way). Events $A$ and it negation $A'$ are disjunct but they are (very) dependent. If one hapens the other does not.
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
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Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
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– user249018
Jan 1 at 0:13
add a comment |
$begingroup$
No, such events are dependent (in extrem way). Events $A$ and it negation $A'$ are disjunct but they are (very) dependent. If one hapens the other does not.
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
$endgroup$
No, such events are dependent (in extrem way). Events $A$ and it negation $A'$ are disjunct but they are (very) dependent. If one hapens the other does not.
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
answered Dec 31 '18 at 23:30
Maria MazurMaria Mazur
48.5k1260121
48.5k1260121
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Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
$endgroup$
– user249018
Jan 1 at 0:13
add a comment |
$begingroup$
Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
$endgroup$
– user249018
Jan 1 at 0:13
$begingroup$
Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
$endgroup$
– user249018
Jan 1 at 0:13
$begingroup$
Thanks. I dont understand what is the negation event in my case. Let say the Event $A$ has a certain outcome. What do you mean by saying that the other does not happen ?
$endgroup$
– user249018
Jan 1 at 0:13
add a comment |
$begingroup$
In a suitable model of reality in the form of a probability space $(Omega,mathcal A,P)$ the described events $A$ and $B$ will be both subsets of the $sigma$-algebra that is in use.
Further the probability measure that is in use will - if it indeed models reality on a suitable way - give: $$P(Acap B)=P(A)P(B)$$
The event $Acap B$ will - if $P(A)$ and $P(B)$ are both positive - not be empty.
If e.g. $(Omega_1,mathcal A_1,P_1)$ models the coin-flipping and $(Omega_2,mathcal A_2,P_2)$ models the weather forcasting then go for the product of both probability spaces.
answered Jan 1 at 11:36
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In a suitable model of reality in the form of a probability space $(Omega,mathcal A,P)$ the described events $A$ and $B$ will be both subsets of the $sigma$-algebra that is in use.
Further the probability measure that is in use will - if it indeed models reality on a suitable way - give: $$P(Acap B)=P(A)P(B)$$
The event $Acap B$ will - if $P(A)$ and $P(B)$ are both positive - not be empty.
If e.g. $(Omega_1,mathcal A_1,P_1)$ models the coin-flipping and $(Omega_2,mathcal A_2,P_2)$ models the weather forcasting then go for the product of both probability spaces.
answered Jan 1 at 11:36
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In a suitable model of reality in the form of a probability space $(Omega,mathcal A,P)$ the described events $A$ and $B$ will be both subsets of the $sigma$-algebra that is in use.
Further the probability measure that is in use will - if it indeed models reality on a suitable way - give: $$P(Acap B)=P(A)P(B)$$
The event $Acap B$ will - if $P(A)$ and $P(B)$ are both positive - not be empty.
If e.g. $(Omega_1,mathcal A_1,P_1)$ models the coin-flipping and $(Omega_2,mathcal A_2,P_2)$ models the weather forcasting then go for the product of both probability spaces.
answered Jan 1 at 11:36
drhabdrhab
103k545136
$endgroup$
In a suitable model of reality in the form of a probability space $(Omega,mathcal A,P)$ the described events $A$ and $B$ will be both subsets of the $sigma$-algebra that is in use.
Further the probability measure that is in use will - if it indeed models reality on a suitable way - give: $$P(Acap B)=P(A)P(B)$$
The event $Acap B$ will - if $P(A)$ and $P(B)$ are both positive - not be empty.
If e.g. $(Omega_1,mathcal A_1,P_1)$ models the coin-flipping and $(Omega_2,mathcal A_2,P_2)$ models the weather forcasting then go for the product of both probability spaces.
answered Jan 1 at 11:36
drhabdrhab
103k545136
answered Jan 1 at 11:36
drhabdrhab
103k545136
answered Jan 1 at 11:36
drhabdrhab
103k545136
answered Jan 1 at 11:36
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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Yes, I'm not sure I understand the confusion here.
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– Don Thousand
Dec 31 '18 at 23:30
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@Jyrki Lahtonen. I wrote $P(A), P(B)$ are non zero and I repeted it again the text. I read your answer in the ''duplicate'' Question where you are comparing the outcome of two rolling dice. Why are the events $A, B$ as stated in my question not Independent but rather exclusive events?
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– user249018
Jan 1 at 0:00
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Sorry about the misread. New Year's Eve party fog. Why do you think that the event of both the coinflip giving tails and the weather forecast saying sunny is empty? Imagine that every morning you flip a coin and read the forecast. Sooner rather than later you get "tails" and "sunny" on the same day. That's not an empty set!
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– Jyrki Lahtonen
Jan 1 at 5:07
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To continue the analogy with dice. Here your coinflip has the role of rolling a red die and the weather forecast is rolling a green die. The outcomes of the two rolls are independent. And $P(R5 text{and} G6)=P(R5)cdot P(G6)$. On the other hand, the events red $5$ and red $6$ are mutually exclusive and have an empty intersection.
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– Jyrki Lahtonen
Jan 1 at 5:14
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Or yet in other words, you were thinking about the intersection in the wrong space. Your universe $Omega$ consists of pairs $$Omega={(a,b)mid text{$a$ is the result of a coinflip}, text{$b$ is the weather forecast}}.$$ So $Omega$ has elements like (heads, sunny), (tails, sunny), (heads, rainy), (tails, rainy) etc.
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– Jyrki Lahtonen
Jan 1 at 5:19