Shouldn't the joint probability of 2 independent events be equal to zero?
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If the joint probability is the intersection of 2 events, then shouldn't the joint probability of 2 independent events be zero since they don't intersect at all? I'm confused.
probability joint-distribution
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up vote
26
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If the joint probability is the intersection of 2 events, then shouldn't the joint probability of 2 independent events be zero since they don't intersect at all? I'm confused.
probability joint-distribution
17
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
2
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
2
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59
add a comment |
up vote
26
down vote
favorite
up vote
26
down vote
favorite
If the joint probability is the intersection of 2 events, then shouldn't the joint probability of 2 independent events be zero since they don't intersect at all? I'm confused.
probability joint-distribution
If the joint probability is the intersection of 2 events, then shouldn't the joint probability of 2 independent events be zero since they don't intersect at all? I'm confused.
probability joint-distribution
probability joint-distribution
asked Dec 7 at 8:35
gaston
16625
16625
17
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
2
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
2
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59
add a comment |
17
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
2
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
2
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59
17
17
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
2
2
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
2
2
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59
add a comment |
3 Answers
3
active
oldest
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up vote
52
down vote
There is a difference between
- independent events: $mathbb P(A cap B) =mathbb P(A),mathbb P(B)$, i.e. $mathbb P(A mid B)= mathbb P(A)$ so knowing one happened gives no information about whether the other happened
- mutually disjoint events: $mathbb P(A cap B) = 0$, i.e. $mathbb P(A mid B)= 0$ so knowing one happened means the other did not happen
You asked for a picture. This might help:
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
|
show 9 more comments
up vote
13
down vote
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The intersect of such events is always 0.
independent events: Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. Let's build on that example:
We roll two dice, a red and a blue. The probability of getting a 1 on the red is given by P(red = 1) = 1/6, and the probability of getting a 1 on the white is given by P(white = 1) = 1/6. It is possible to get their intersect (i.e. both get 1) simply by multiplying them, since they are independent. P(red = 1) x P(white = 1) = 1/6 x 1/6 = 1/36 != 0.
In simple words 1/6 of the time the red die is a 1, and 1/6 of those times the white die is 1. To illustrate:
add a comment |
up vote
2
down vote
The confusion of the OP lies on the notions of disjoint events and independent events.
One simple and intuitive description of independence is:
A and B are independent if knowing that A happened gives you no
information about whether or not B happened.
Or in other words,
A and B are independent if knowing that A happened does not change the
probability that B happened.
If A and B are disjoint then knowing that A happened is a game changer! Now you would be certain that B did not happen! And so they are not independent.
The only way independence and "disjointedness" in this example are the same is when B is the empty set (which has probability 0). In this case A happening does not inform anything on B
No pictures but at least some intuition
add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
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up vote
52
down vote
There is a difference between
- independent events: $mathbb P(A cap B) =mathbb P(A),mathbb P(B)$, i.e. $mathbb P(A mid B)= mathbb P(A)$ so knowing one happened gives no information about whether the other happened
- mutually disjoint events: $mathbb P(A cap B) = 0$, i.e. $mathbb P(A mid B)= 0$ so knowing one happened means the other did not happen
You asked for a picture. This might help:
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
|
show 9 more comments
up vote
52
down vote
There is a difference between
- independent events: $mathbb P(A cap B) =mathbb P(A),mathbb P(B)$, i.e. $mathbb P(A mid B)= mathbb P(A)$ so knowing one happened gives no information about whether the other happened
- mutually disjoint events: $mathbb P(A cap B) = 0$, i.e. $mathbb P(A mid B)= 0$ so knowing one happened means the other did not happen
You asked for a picture. This might help:
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
|
show 9 more comments
up vote
52
down vote
up vote
52
down vote
There is a difference between
- independent events: $mathbb P(A cap B) =mathbb P(A),mathbb P(B)$, i.e. $mathbb P(A mid B)= mathbb P(A)$ so knowing one happened gives no information about whether the other happened
- mutually disjoint events: $mathbb P(A cap B) = 0$, i.e. $mathbb P(A mid B)= 0$ so knowing one happened means the other did not happen
You asked for a picture. This might help:
There is a difference between
- independent events: $mathbb P(A cap B) =mathbb P(A),mathbb P(B)$, i.e. $mathbb P(A mid B)= mathbb P(A)$ so knowing one happened gives no information about whether the other happened
- mutually disjoint events: $mathbb P(A cap B) = 0$, i.e. $mathbb P(A mid B)= 0$ so knowing one happened means the other did not happen
You asked for a picture. This might help:
edited Dec 8 at 1:26
user985366
1032
1032
answered Dec 7 at 8:41
Henry
18.5k13466
18.5k13466
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
|
show 9 more comments
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
7
7
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
Is there a reason you wrote "almost" in the second bullet point? Is that one of those "possible with probability zero" things? I would think it is by definition impossible (such as the probability of heads and the probability of tails), then why write "almost certainly" rather than "certainly"? I suppose this is the probabilistic interpretation.
– gerrit
Dec 7 at 10:51
3
3
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
@Barranka I get that, but that doesn't look like what is drawn in the picture on right. The joint probability of a uniformly drawn random number in [0, 1] being both smaller than 0.4 and larger than 0.6 is not only zero, it is also completely impossible. Isn't that what the wide band in the right figure illustrates? Or am I misreading the figure?
– gerrit
Dec 7 at 13:13
2
2
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
@Barranka I could throw the coin so fast that it escapes earth's gravitational pull. I would venture P(HEADS)=0.499..., P(TAILS)=0.499..., 0<P(LAND ON SIDE)<0.000000000001, and 0<P(ESCAPE VELOCITY)<0.0000000000001. Strictly speaking if the probability of an event is zero, then it can not happen.
– emory
Dec 7 at 13:24
4
4
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
I'm no expert, but even after your last comment I agree with @gerrit: Heads and Tails are disjoint. It's possible to get Not heads and not tails, but it's impossible to get heads and tails. Thus knowing heads happened means that tails could not possibly have happened - no "almost" about it. I could be wrong on my terminology, but if so please explain patiently as I'm not the only one missing it
– Chris H
Dec 7 at 13:53
2
2
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
@Braanka Your coin example is a poor one, as presumably landing on a side has non-zero probability, and if you say that it has zero probability, well, now you're pretty much just begging the question.
– Acccumulation
Dec 7 at 19:45
|
show 9 more comments
up vote
13
down vote
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The intersect of such events is always 0.
independent events: Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. Let's build on that example:
We roll two dice, a red and a blue. The probability of getting a 1 on the red is given by P(red = 1) = 1/6, and the probability of getting a 1 on the white is given by P(white = 1) = 1/6. It is possible to get their intersect (i.e. both get 1) simply by multiplying them, since they are independent. P(red = 1) x P(white = 1) = 1/6 x 1/6 = 1/36 != 0.
In simple words 1/6 of the time the red die is a 1, and 1/6 of those times the white die is 1. To illustrate:
add a comment |
up vote
13
down vote
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The intersect of such events is always 0.
independent events: Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. Let's build on that example:
We roll two dice, a red and a blue. The probability of getting a 1 on the red is given by P(red = 1) = 1/6, and the probability of getting a 1 on the white is given by P(white = 1) = 1/6. It is possible to get their intersect (i.e. both get 1) simply by multiplying them, since they are independent. P(red = 1) x P(white = 1) = 1/6 x 1/6 = 1/36 != 0.
In simple words 1/6 of the time the red die is a 1, and 1/6 of those times the white die is 1. To illustrate:
add a comment |
up vote
13
down vote
up vote
13
down vote
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The intersect of such events is always 0.
independent events: Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. Let's build on that example:
We roll two dice, a red and a blue. The probability of getting a 1 on the red is given by P(red = 1) = 1/6, and the probability of getting a 1 on the white is given by P(white = 1) = 1/6. It is possible to get their intersect (i.e. both get 1) simply by multiplying them, since they are independent. P(red = 1) x P(white = 1) = 1/6 x 1/6 = 1/36 != 0.
In simple words 1/6 of the time the red die is a 1, and 1/6 of those times the white die is 1. To illustrate:
What I understood from your question, is that you might have confused independent events with disjoint events.
disjoint events: Two events are called disjoint or mutually exclusive if they cannot both happen. For instance, if we roll a die, the outcomes 1 and 2 are disjoint since they cannot both occur. On the other hand, the outcomes 1 and “rolling an odd number” are not disjoint since both occur if the outcome of the roll is a 1. The intersect of such events is always 0.
independent events: Two events are independent if knowing the outcome of one provides no useful information about the outcome of the other. For instance, when we roll two dice, the outcome of each is an independent event – knowing the outcome of one roll does not help determining the outcome of the other. Let's build on that example:
We roll two dice, a red and a blue. The probability of getting a 1 on the red is given by P(red = 1) = 1/6, and the probability of getting a 1 on the white is given by P(white = 1) = 1/6. It is possible to get their intersect (i.e. both get 1) simply by multiplying them, since they are independent. P(red = 1) x P(white = 1) = 1/6 x 1/6 = 1/36 != 0.
In simple words 1/6 of the time the red die is a 1, and 1/6 of those times the white die is 1. To illustrate:
edited Dec 7 at 10:05
answered Dec 7 at 9:55
Umair Rafique
1314
1314
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The confusion of the OP lies on the notions of disjoint events and independent events.
One simple and intuitive description of independence is:
A and B are independent if knowing that A happened gives you no
information about whether or not B happened.
Or in other words,
A and B are independent if knowing that A happened does not change the
probability that B happened.
If A and B are disjoint then knowing that A happened is a game changer! Now you would be certain that B did not happen! And so they are not independent.
The only way independence and "disjointedness" in this example are the same is when B is the empty set (which has probability 0). In this case A happening does not inform anything on B
No pictures but at least some intuition
add a comment |
up vote
2
down vote
The confusion of the OP lies on the notions of disjoint events and independent events.
One simple and intuitive description of independence is:
A and B are independent if knowing that A happened gives you no
information about whether or not B happened.
Or in other words,
A and B are independent if knowing that A happened does not change the
probability that B happened.
If A and B are disjoint then knowing that A happened is a game changer! Now you would be certain that B did not happen! And so they are not independent.
The only way independence and "disjointedness" in this example are the same is when B is the empty set (which has probability 0). In this case A happening does not inform anything on B
No pictures but at least some intuition
add a comment |
up vote
2
down vote
up vote
2
down vote
The confusion of the OP lies on the notions of disjoint events and independent events.
One simple and intuitive description of independence is:
A and B are independent if knowing that A happened gives you no
information about whether or not B happened.
Or in other words,
A and B are independent if knowing that A happened does not change the
probability that B happened.
If A and B are disjoint then knowing that A happened is a game changer! Now you would be certain that B did not happen! And so they are not independent.
The only way independence and "disjointedness" in this example are the same is when B is the empty set (which has probability 0). In this case A happening does not inform anything on B
No pictures but at least some intuition
The confusion of the OP lies on the notions of disjoint events and independent events.
One simple and intuitive description of independence is:
A and B are independent if knowing that A happened gives you no
information about whether or not B happened.
Or in other words,
A and B are independent if knowing that A happened does not change the
probability that B happened.
If A and B are disjoint then knowing that A happened is a game changer! Now you would be certain that B did not happen! And so they are not independent.
The only way independence and "disjointedness" in this example are the same is when B is the empty set (which has probability 0). In this case A happening does not inform anything on B
No pictures but at least some intuition
answered Dec 7 at 15:41
gota
1437
1437
add a comment |
add a comment |
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17
The probability that I watch on a given day TV is 1/2. The probability that it rains on a given day is 1/2. These are independent events. What is the probability of me watching TV on a rainy day?
– user1936752
Dec 7 at 18:18
2
@user1936752 Strictly speaking, your example events are not independent for most people (e.g., they might be more willing to spend time outdoors when it does not rain)
– Hagen von Eitzen
Dec 9 at 12:41
@HagenvonEitzen OK, good point. Change rainy day to eat chocolate.
– Rui Barradas
Dec 10 at 10:44
2
@Gaston: Don't confuse "independent" with "mutually exclusive". Independent events are completely unrelated to each other, whereas mutually exclusive events are inherently related. For example, suppose I flip two coins: whether I get heads on Coin 1 is unaffected by the result of Coin 2, but it's inherently connected to whether I get tails on Coin 1! =)
– jdmc
Dec 10 at 17:45
This video here and this other one will be helpful in understanding these concepts.
– Learn_Share
Dec 11 at 22:59