Why independent events are never mutually exclusive?
up vote
2
down vote
favorite
I am a novice at probability and I came across this statement in my maths book.
Two independent events are never mutually exclusive. I am not getting why it is so. Can anyone explain using a ven diagram?
probability
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
add a comment |
up vote
2
down vote
favorite
I am a novice at probability and I came across this statement in my maths book.
Two independent events are never mutually exclusive. I am not getting why it is so. Can anyone explain using a ven diagram?
probability
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am a novice at probability and I came across this statement in my maths book.
Two independent events are never mutually exclusive. I am not getting why it is so. Can anyone explain using a ven diagram?
probability
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
I am a novice at probability and I came across this statement in my maths book.
Two independent events are never mutually exclusive. I am not getting why it is so. Can anyone explain using a ven diagram?
probability
probability
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
asked Jul 29 '17 at 4:12
Deepeshkumar
1275
1275
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(Acap B)=0$. But if they are independent then $P(Acap B)=P(A)P(B)>0$. Contradicion.
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
add a comment |
up vote
2
down vote
It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A cap lnot B$ region.
Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(Acap B)=0$. But if they are independent then $P(Acap B)=P(A)P(B)>0$. Contradicion.
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
add a comment |
up vote
2
down vote
accepted
Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(Acap B)=0$. But if they are independent then $P(Acap B)=P(A)P(B)>0$. Contradicion.
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(Acap B)=0$. But if they are independent then $P(Acap B)=P(A)P(B)>0$. Contradicion.
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(Acap B)=0$. But if they are independent then $P(Acap B)=P(A)P(B)>0$. Contradicion.
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
answered Jul 29 '17 at 4:23
MAN-MADE
3,5731931
3,5731931
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
add a comment |
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false.
– hardmath
Jul 29 '17 at 4:32
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
@hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold.
– MAN-MADE
Jul 29 '17 at 4:33
add a comment |
up vote
2
down vote
It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A cap lnot B$ region.
Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
|
show 1 more comment
up vote
2
down vote
It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A cap lnot B$ region.
Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
|
show 1 more comment
up vote
2
down vote
up vote
2
down vote
It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A cap lnot B$ region.
Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A cap lnot B$ region.
Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
answered Jul 29 '17 at 4:24
Ross Millikan
290k23195368
290k23195368
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
|
show 1 more comment
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
Sir, what is $A cap lnot B$ (not familiar with the notation!)
– MAN-MADE
Jul 29 '17 at 4:29
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
It is the intersection of $A$ and $lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A cap B$. The answer you accepted assumes $P(A) gt 0, P(B) gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false.
– Ross Millikan
Jul 29 '17 at 4:32
1
1
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
Also $neg B$ is the complement of $B$.
– Graham Kemp
Jul 29 '17 at 4:35
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution!
– MAN-MADE
Jul 29 '17 at 4:39
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
@GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that...
– MAN-MADE
Jul 29 '17 at 4:46
|
show 1 more comment
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