Which probability is affected by dependent events- P(A) or P(A|C)?Intuitive Understanding of Dependent Events in ProbabilityWhat is the meaning of “independent events ” and how can we logically conclude independence of two events in probability?Probability of independent & mutually exclusive eventsI can't grasp how these events are independent?Independence intuitionProbability of joint events which are dependentDependent eventsAre all disjoint events dependent?Independent Events Conceptual Meaning - probability theoryDetermining whether two events are independent or dependent.
If human space travel is limited by the G force vulnerability, is there a way to counter G forces?
Arrow those variables!
Can I make popcorn with any corn?
infared filters v nd
How to format long polynomial?
Can I ask the recruiters in my resume to put the reason why I am rejected?
Does detail obscure or enhance action?
How can I prevent hyper evolved versions of regular creatures from wiping out their cousins?
Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?
What is the word for reserving something for yourself before others do?
Why is Minecraft giving an OpenGL error?
Cross compiling for RPi - error while loading shared libraries
RSA: Danger of using p to create q
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?
Horror movie about a virus at the prom; beginning and end are stylized as a cartoon
Codimension of non-flat locus
How to move a thin line with the black arrow in Illustrator?
How do I deal with an unproductive colleague in a small company?
What's the point of deactivating Num Lock on login screens?
Convert two switches to a dual stack, and add outlet - possible here?
Java Casting: Java 11 throws LambdaConversionException while 1.8 does not
How does quantile regression compare to logistic regression with the variable split at the quantile?
Malformed Address '10.10.21.08/24', must be X.X.X.X/NN or
Is it legal for company to use my work email to pretend I still work there?
Which probability is affected by dependent events- P(A) or P(A|C)?
Intuitive Understanding of Dependent Events in ProbabilityWhat is the meaning of “independent events ” and how can we logically conclude independence of two events in probability?Probability of independent & mutually exclusive eventsI can't grasp how these events are independent?Independence intuitionProbability of joint events which are dependentDependent eventsAre all disjoint events dependent?Independent Events Conceptual Meaning - probability theoryDetermining whether two events are independent or dependent.
$begingroup$
Def1. INDEPENDENT EVENTS: two events A and B are independent if occurrence of B does not affect the probability of occurrence of A (and vice versa). [A-ind-B in short]
Def2. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of occurrence of the A (and vice versa). [A-dep-C in short]
Let us have A-ind-B and A-dep-C.
So we get:
N1. B doesn't affect P(A) by above definition.
N2. C affects P(A) by above definition.
But in reality, even if C occurs, it has nothing to do with probability of occurrence of A, P(A).
In other words even though A-dep-C , C still doesn't affect P(A).
So isn't A and C independent by Def1 (because C doesn't affect P(A) )? But A and C are NOT independent, they are dependent (we have assigned them to be dependent).
Let's change the definition of Dependent events:
Def3. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of 'occurrence of A such that C has occurred' (i.e. P(A|C) ) (and vice versa).
In other words, if A-dep-C , then occurrence of C affects P(A|C) (and not P(A) )
So if we compare Def2 and Def3 , this means "probability of occurrence of A" (as in Def2) actually refers to P(A|C) and NOT to P(A). That is P(A|C) is affected, and not P(A).
1. So which one is affected- P(A) or P(A|C) ?
2. And which one should be definition of dependent events- Def2 or Def3 ?
probability conditional-probability independence
asked Mar 29 at 10:34
Ane SaAne Sa
62
$endgroup$
|
show 1 more comment
$begingroup$
Def1. INDEPENDENT EVENTS: two events A and B are independent if occurrence of B does not affect the probability of occurrence of A (and vice versa). [A-ind-B in short]
Def2. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of occurrence of the A (and vice versa). [A-dep-C in short]
Let us have A-ind-B and A-dep-C.
So we get:
N1. B doesn't affect P(A) by above definition.
N2. C affects P(A) by above definition.
But in reality, even if C occurs, it has nothing to do with probability of occurrence of A, P(A).
In other words even though A-dep-C , C still doesn't affect P(A).
So isn't A and C independent by Def1 (because C doesn't affect P(A) )? But A and C are NOT independent, they are dependent (we have assigned them to be dependent).
Let's change the definition of Dependent events:
Def3. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of 'occurrence of A such that C has occurred' (i.e. P(A|C) ) (and vice versa).
In other words, if A-dep-C , then occurrence of C affects P(A|C) (and not P(A) )
So if we compare Def2 and Def3 , this means "probability of occurrence of A" (as in Def2) actually refers to P(A|C) and NOT to P(A). That is P(A|C) is affected, and not P(A).
1. So which one is affected- P(A) or P(A|C) ?
2. And which one should be definition of dependent events- Def2 or Def3 ?
probability conditional-probability independence
asked Mar 29 at 10:34
Ane SaAne Sa
62
$endgroup$
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54
|
show 1 more comment
$begingroup$
Def1. INDEPENDENT EVENTS: two events A and B are independent if occurrence of B does not affect the probability of occurrence of A (and vice versa). [A-ind-B in short]
Def2. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of occurrence of the A (and vice versa). [A-dep-C in short]
Let us have A-ind-B and A-dep-C.
So we get:
N1. B doesn't affect P(A) by above definition.
N2. C affects P(A) by above definition.
But in reality, even if C occurs, it has nothing to do with probability of occurrence of A, P(A).
In other words even though A-dep-C , C still doesn't affect P(A).
So isn't A and C independent by Def1 (because C doesn't affect P(A) )? But A and C are NOT independent, they are dependent (we have assigned them to be dependent).
Let's change the definition of Dependent events:
Def3. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of 'occurrence of A such that C has occurred' (i.e. P(A|C) ) (and vice versa).
In other words, if A-dep-C , then occurrence of C affects P(A|C) (and not P(A) )
So if we compare Def2 and Def3 , this means "probability of occurrence of A" (as in Def2) actually refers to P(A|C) and NOT to P(A). That is P(A|C) is affected, and not P(A).
1. So which one is affected- P(A) or P(A|C) ?
2. And which one should be definition of dependent events- Def2 or Def3 ?
probability conditional-probability independence
asked Mar 29 at 10:34
Ane SaAne Sa
62
$endgroup$
Def1. INDEPENDENT EVENTS: two events A and B are independent if occurrence of B does not affect the probability of occurrence of A (and vice versa). [A-ind-B in short]
Def2. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of occurrence of the A (and vice versa). [A-dep-C in short]
Let us have A-ind-B and A-dep-C.
So we get:
N1. B doesn't affect P(A) by above definition.
N2. C affects P(A) by above definition.
But in reality, even if C occurs, it has nothing to do with probability of occurrence of A, P(A).
In other words even though A-dep-C , C still doesn't affect P(A).
So isn't A and C independent by Def1 (because C doesn't affect P(A) )? But A and C are NOT independent, they are dependent (we have assigned them to be dependent).
Let's change the definition of Dependent events:
Def3. DEPENDENT EVENTS: two events A and C are dependent if occurrence of C affects the probability of 'occurrence of A such that C has occurred' (i.e. P(A|C) ) (and vice versa).
In other words, if A-dep-C , then occurrence of C affects P(A|C) (and not P(A) )
So if we compare Def2 and Def3 , this means "probability of occurrence of A" (as in Def2) actually refers to P(A|C) and NOT to P(A). That is P(A|C) is affected, and not P(A).
1. So which one is affected- P(A) or P(A|C) ?
2. And which one should be definition of dependent events- Def2 or Def3 ?
probability conditional-probability independence
probability conditional-probability independence
asked Mar 29 at 10:34
Ane SaAne Sa
62
asked Mar 29 at 10:34
Ane SaAne Sa
62
edited Mar 29 at 21:41
Ane Sa
asked Mar 29 at 10:34
Ane SaAne Sa
62
asked Mar 29 at 10:34
Ane SaAne Sa
62
asked Mar 29 at 10:34
Ane SaAne Sa
62
62
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54
|
show 1 more comment
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54
|
show 1 more comment
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166985%2fwhich-probability-is-affected-by-dependent-events-pa-or-pac%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166985%2fwhich-probability-is-affected-by-dependent-events-pa-or-pac%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This is not clear. It is not generally true that "in reality, even if C occurs, it has nothing to do with probability of occurrence of $A$, $P(A)$. In other words even though A-dep-C , C still doesn't affect P(A)." Say you are tossing a fair coin, $A$ is the event that it comes up $H$ and $C$ is the event that it comes up $T$. Then $P(A)=.5$ but $P(A,|,C)=0$. Informally, knowing that $C$ has occurred excludes the chance that $A$ also occurred.
$endgroup$
– lulu
Mar 29 at 11:01
$begingroup$
The wiki article is, I think, quite clear on the definition.
$endgroup$
– lulu
Mar 29 at 11:04
$begingroup$
Keep in mind: informal definitions are just that, informal. They serve a purpose in that they communicate the intent of the definition in simple terms. Usually they are not good mathematical definitions. They use imprecise, ill-defined terms. In order to be useful mathematically they must be accompanied with a formal definition. The wiki article I linked to does that...it starts with the "ordinary language" definition which, as you remark, is imprecise....
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
...But then it supplies the formal definition...$A,B$ are independent if $P(Acap B)=P(A)times P(B)$. They then show that this is equivalent to $P(A)=P(A,|,B)$ which, I'd say, is the best mathematical translation of the informal definition.
$endgroup$
– lulu
Mar 29 at 11:10
$begingroup$
@lulu thanks for your help. The word definition or informal definition really confused me while studying independent events. You explaination really helped me and you made me understand the concept better. Thanks for help. $_/backslash_$
$endgroup$
– Ane Sa
Mar 29 at 21:54