Mutually independent $P(A ∩ B ∩ C) ne P(A ∩ B∩ A ∩ C)$












1












$begingroup$


I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$



Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Who said $Acap B$ and $Acap C$ are independent?
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 15:58










  • $begingroup$
    @RushabhMehta That's given.
    $endgroup$
    – Paprika
    Dec 17 '18 at 16:01






  • 1




    $begingroup$
    $P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:29








  • 1




    $begingroup$
    Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:31






  • 1




    $begingroup$
    I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:38


















1












$begingroup$


I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$



Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Who said $Acap B$ and $Acap C$ are independent?
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 15:58










  • $begingroup$
    @RushabhMehta That's given.
    $endgroup$
    – Paprika
    Dec 17 '18 at 16:01






  • 1




    $begingroup$
    $P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:29








  • 1




    $begingroup$
    Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:31






  • 1




    $begingroup$
    I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:38
















1












1








1


0



$begingroup$


I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$



Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.










share|cite|improve this question











$endgroup$




I have events $A$, $B$, $C$ which are mutually independent. Let's say $A∩B$ and $A∩C$ are also independent. Then $$P(A ∩ B ∩ C) = P(A)P(B)P(C)$$ but
$$P(A ∩ B ∩ C) = P((A ∩ B) ∩ (A ∩ C)) = P(A ∩ B)P(A ∩ C) = P(A)P(B)P(A)P(C)$$
$$P(A)P(B)P(C) = P(A)P(B)P(A)P(C)$$



Why is it so? I guess that $A = Omega$ or $emptyset$? How then this conclusion can be drawn from given properties? Would really appreciate any input.







probability elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 15:57









Tianlalu

3,09421138




3,09421138










asked Dec 17 '18 at 15:42









PaprikaPaprika

113




113








  • 1




    $begingroup$
    Who said $Acap B$ and $Acap C$ are independent?
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 15:58










  • $begingroup$
    @RushabhMehta That's given.
    $endgroup$
    – Paprika
    Dec 17 '18 at 16:01






  • 1




    $begingroup$
    $P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:29








  • 1




    $begingroup$
    Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:31






  • 1




    $begingroup$
    I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:38
















  • 1




    $begingroup$
    Who said $Acap B$ and $Acap C$ are independent?
    $endgroup$
    – Don Thousand
    Dec 17 '18 at 15:58










  • $begingroup$
    @RushabhMehta That's given.
    $endgroup$
    – Paprika
    Dec 17 '18 at 16:01






  • 1




    $begingroup$
    $P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:29








  • 1




    $begingroup$
    Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:31






  • 1




    $begingroup$
    I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
    $endgroup$
    – user593746
    Dec 17 '18 at 16:38










1




1




$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58




$begingroup$
Who said $Acap B$ and $Acap C$ are independent?
$endgroup$
– Don Thousand
Dec 17 '18 at 15:58












$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01




$begingroup$
@RushabhMehta That's given.
$endgroup$
– Paprika
Dec 17 '18 at 16:01




1




1




$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29






$begingroup$
$P(A)=0$ or $P(A)=1$ does not imply that $A=varnothing$ or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:29






1




1




$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31




$begingroup$
Take $Omega=Bbb{R}$ for example, say, with the standard normal distribution. Then, $A={0}$ is not empty but $P(A)=0$. Similarly, $A=Bbb{R}setminus{0}$ is not $Omega$ but $P(A)=1$.
$endgroup$
– user593746
Dec 17 '18 at 16:31




1




1




$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38






$begingroup$
I don't know. What is $Omega$ here? What is the probability distribution on $Omega$? If we don't know these two pieces of information, I'm afraid you cannot conclude $A=varnothing $or $A=Omega$.
$endgroup$
– user593746
Dec 17 '18 at 16:38












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044097%2fmutually-independent-pa-%25e2%2588%25a9-b-%25e2%2588%25a9-c-ne-pa-%25e2%2588%25a9-b%25e2%2588%25a9-a-%25e2%2588%25a9-c%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044097%2fmutually-independent-pa-%25e2%2588%25a9-b-%25e2%2588%25a9-c-ne-pa-%25e2%2588%25a9-b%25e2%2588%25a9-a-%25e2%2588%25a9-c%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Plaza Victoria

Puebla de Zaragoza

Musa