Independent events in set theory [closed]

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If $A_1$, $A_2$, $A_3$ are mutually independent events. How do I show that $A_1 cup A_2$ and $A_3$ are also independent? I am able to prove that $A_1$, $A_2$, $A_3$ using this $$P(A_1 cap A_2 cap A_3) = P(A_1) cap P(A_2) cap P(A_3)$$ But showing that $A_1 cup A_2$ and $A_3$ is what I am struggling to prove.




statistics statistical-inference




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closed as off-topic by Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom Aug 25 at 13:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
    – Henrik
    Aug 25 at 8:17






  • 1




    @SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
    – mathcounterexamples.net
    Aug 25 at 8:23











  • @mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
    – Sujit Bhattacharyya
    Aug 25 at 12:35














up vote
1
down vote

favorite












If $A_1$, $A_2$, $A_3$ are mutually independent events. How do I show that $A_1 cup A_2$ and $A_3$ are also independent? I am able to prove that $A_1$, $A_2$, $A_3$ using this $$P(A_1 cap A_2 cap A_3) = P(A_1) cap P(A_2) cap P(A_3)$$ But showing that $A_1 cup A_2$ and $A_3$ is what I am struggling to prove.




statistics statistical-inference




share|cite|improve this question














closed as off-topic by Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom Aug 25 at 13:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
    – Henrik
    Aug 25 at 8:17






  • 1




    @SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
    – mathcounterexamples.net
    Aug 25 at 8:23











  • @mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
    – Sujit Bhattacharyya
    Aug 25 at 12:35












up vote
1
down vote

favorite









up vote
1
down vote

favorite











If $A_1$, $A_2$, $A_3$ are mutually independent events. How do I show that $A_1 cup A_2$ and $A_3$ are also independent? I am able to prove that $A_1$, $A_2$, $A_3$ using this $$P(A_1 cap A_2 cap A_3) = P(A_1) cap P(A_2) cap P(A_3)$$ But showing that $A_1 cup A_2$ and $A_3$ is what I am struggling to prove.




statistics statistical-inference




share|cite|improve this question














If $A_1$, $A_2$, $A_3$ are mutually independent events. How do I show that $A_1 cup A_2$ and $A_3$ are also independent? I am able to prove that $A_1$, $A_2$, $A_3$ using this $$P(A_1 cap A_2 cap A_3) = P(A_1) cap P(A_2) cap P(A_3)$$ But showing that $A_1 cup A_2$ and $A_3$ is what I am struggling to prove.





statistics statistical-inference





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 15:38

























asked Aug 25 at 8:12









Lady

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closed as off-topic by Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom Aug 25 at 13:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom Aug 25 at 13:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henrik, Jendrik Stelzner, Brahadeesh, user91500, StubbornAtom
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
    – Henrik
    Aug 25 at 8:17






  • 1




    @SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
    – mathcounterexamples.net
    Aug 25 at 8:23











  • @mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
    – Sujit Bhattacharyya
    Aug 25 at 12:35












  • 2




    You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
    – Henrik
    Aug 25 at 8:17






  • 1




    @SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
    – mathcounterexamples.net
    Aug 25 at 8:23











  • @mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
    – Sujit Bhattacharyya
    Aug 25 at 12:35







2




2




You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
– Henrik
Aug 25 at 8:17




You use the definition. And if you want better help, you include some of your own work/thouhgts in the question so we can see where you struggle and offer help with that instead of just providing a solution.
– Henrik
Aug 25 at 8:17




1




1




@SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
– mathcounterexamples.net
Aug 25 at 8:23





@SujitBhattacharyya If I may... You’re saying trivial, paraphrasing what is requested without bringing a real proof.
– mathcounterexamples.net
Aug 25 at 8:23













@mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
– Sujit Bhattacharyya
Aug 25 at 12:35




@mathcounterexamples.net I apologize for that and I agree with Henrik, the definition is sufficient.
– Sujit Bhattacharyya
Aug 25 at 12:35















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