Mixing
Proof Concerning the Union of Events A,B,C [duplicate]
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This question already has an answer here:
Probability of the union of $3$ events?
3 answers
${P(A}{cup}{B}{cup}{C)}={P(A)+P(B)+P(C)}-{P(A{cap}B)}-{P(A{cap}C)}-{P(B{cap}C)}+{P(A{cap}B{cap}C)}$
Proof:
${P(A{cup}B{cup}C)}={P(A{cup}(B{cup}C))}$ Associativity
${P(A{cup}(B{cup}C))}={P(A)+P(B{cup}C)-P(A{cap}(B{cup}C))}$ Well-known theorem
${P(A{cap}(B{cup}C)=P(A{cap}B){cup}P(A{cap}C)}$ Distributive
Now I should apply the well-known theorem again.
${P(A{cap}B){cup}P(A{cap}C)=P(A{cap}B)+P(A{cap}C)-P((A{cap}B){cap}P(A{cap}C))}$
I'd be finished if I could show this:
${P((A{cap}B){cap}P(A{cap}C))}=P(A{cap}B{cap}C)$
Which I could easily justify using a Venn Diagram (which is not acceptable). How else could I do this?
EDIT: The duplicate question doesn't help me. The answer says "and since (A∩C)∩(B∩C)=A∩B∩C, the result is proven." but this is not explained and I DONT want to use a Venn Diagram.
probability statistics
asked Jan 21 at 4:36
SydneySydney
234
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marked as duplicate by Lord Shark the Unknown, Shailesh, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 11:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Probability of the union of $3$ events?
3 answers
${P(A}{cup}{B}{cup}{C)}={P(A)+P(B)+P(C)}-{P(A{cap}B)}-{P(A{cap}C)}-{P(B{cap}C)}+{P(A{cap}B{cap}C)}$
Proof:
${P(A{cup}B{cup}C)}={P(A{cup}(B{cup}C))}$ Associativity
${P(A{cup}(B{cup}C))}={P(A)+P(B{cup}C)-P(A{cap}(B{cup}C))}$ Well-known theorem
${P(A{cap}(B{cup}C)=P(A{cap}B){cup}P(A{cap}C)}$ Distributive
Now I should apply the well-known theorem again.
${P(A{cap}B){cup}P(A{cap}C)=P(A{cap}B)+P(A{cap}C)-P((A{cap}B){cap}P(A{cap}C))}$
I'd be finished if I could show this:
${P((A{cap}B){cap}P(A{cap}C))}=P(A{cap}B{cap}C)$
Which I could easily justify using a Venn Diagram (which is not acceptable). How else could I do this?
EDIT: The duplicate question doesn't help me. The answer says "and since (A∩C)∩(B∩C)=A∩B∩C, the result is proven." but this is not explained and I DONT want to use a Venn Diagram.
probability statistics
asked Jan 21 at 4:36
SydneySydney
234
$endgroup$
marked as duplicate by Lord Shark the Unknown, Shailesh, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 11:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Probability of the union of $3$ events?
3 answers
${P(A}{cup}{B}{cup}{C)}={P(A)+P(B)+P(C)}-{P(A{cap}B)}-{P(A{cap}C)}-{P(B{cap}C)}+{P(A{cap}B{cap}C)}$
Proof:
${P(A{cup}B{cup}C)}={P(A{cup}(B{cup}C))}$ Associativity
${P(A{cup}(B{cup}C))}={P(A)+P(B{cup}C)-P(A{cap}(B{cup}C))}$ Well-known theorem
${P(A{cap}(B{cup}C)=P(A{cap}B){cup}P(A{cap}C)}$ Distributive
Now I should apply the well-known theorem again.
${P(A{cap}B){cup}P(A{cap}C)=P(A{cap}B)+P(A{cap}C)-P((A{cap}B){cap}P(A{cap}C))}$
I'd be finished if I could show this:
${P((A{cap}B){cap}P(A{cap}C))}=P(A{cap}B{cap}C)$
Which I could easily justify using a Venn Diagram (which is not acceptable). How else could I do this?
EDIT: The duplicate question doesn't help me. The answer says "and since (A∩C)∩(B∩C)=A∩B∩C, the result is proven." but this is not explained and I DONT want to use a Venn Diagram.
probability statistics
asked Jan 21 at 4:36
SydneySydney
234
$endgroup$
This question already has an answer here:
Probability of the union of $3$ events?
3 answers
${P(A}{cup}{B}{cup}{C)}={P(A)+P(B)+P(C)}-{P(A{cap}B)}-{P(A{cap}C)}-{P(B{cap}C)}+{P(A{cap}B{cap}C)}$
Proof:
${P(A{cup}B{cup}C)}={P(A{cup}(B{cup}C))}$ Associativity
${P(A{cup}(B{cup}C))}={P(A)+P(B{cup}C)-P(A{cap}(B{cup}C))}$ Well-known theorem
${P(A{cap}(B{cup}C)=P(A{cap}B){cup}P(A{cap}C)}$ Distributive
Now I should apply the well-known theorem again.
${P(A{cap}B){cup}P(A{cap}C)=P(A{cap}B)+P(A{cap}C)-P((A{cap}B){cap}P(A{cap}C))}$
I'd be finished if I could show this:
${P((A{cap}B){cap}P(A{cap}C))}=P(A{cap}B{cap}C)$
Which I could easily justify using a Venn Diagram (which is not acceptable). How else could I do this?
EDIT: The duplicate question doesn't help me. The answer says "and since (A∩C)∩(B∩C)=A∩B∩C, the result is proven." but this is not explained and I DONT want to use a Venn Diagram.
This question already has an answer here:
Probability of the union of $3$ events?
3 answers
probability statistics
probability statistics
asked Jan 21 at 4:36
SydneySydney
234
asked Jan 21 at 4:36
SydneySydney
234
edited Jan 21 at 5:30
Sydney
asked Jan 21 at 4:36
SydneySydney
234
asked Jan 21 at 4:36
SydneySydney
234
asked Jan 21 at 4:36
SydneySydney
234
234
marked as duplicate by Lord Shark the Unknown, Shailesh, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 11:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, Shailesh, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 11:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
https://en.wikipedia.org/wiki/Algebra_of_sets
If you indeed want to play around with the commutative laws and associative laws, here we go:
$$ begin{align}
(A cap C) cap (B cap C)
&= A cap (C cap (B cap C)) tag{Associative Law} \
&= A cap ((B cap C) cap C) tag{Commutative Law} \
&= A cap (B cap (C cap C)) tag{Associative Law} \
&= A cap (B cap C) \
end{align}$$
answered Jan 21 at 8:23
BGMBGM
3,835158
$endgroup$
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
https://en.wikipedia.org/wiki/Algebra_of_sets
If you indeed want to play around with the commutative laws and associative laws, here we go:
$$ begin{align}
(A cap C) cap (B cap C)
&= A cap (C cap (B cap C)) tag{Associative Law} \
&= A cap ((B cap C) cap C) tag{Commutative Law} \
&= A cap (B cap (C cap C)) tag{Associative Law} \
&= A cap (B cap C) \
end{align}$$
answered Jan 21 at 8:23
BGMBGM
3,835158
$endgroup$
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
add a comment |
$begingroup$
https://en.wikipedia.org/wiki/Algebra_of_sets
If you indeed want to play around with the commutative laws and associative laws, here we go:
$$ begin{align}
(A cap C) cap (B cap C)
&= A cap (C cap (B cap C)) tag{Associative Law} \
&= A cap ((B cap C) cap C) tag{Commutative Law} \
&= A cap (B cap (C cap C)) tag{Associative Law} \
&= A cap (B cap C) \
end{align}$$
answered Jan 21 at 8:23
BGMBGM
3,835158
$endgroup$
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
add a comment |
$begingroup$
https://en.wikipedia.org/wiki/Algebra_of_sets
If you indeed want to play around with the commutative laws and associative laws, here we go:
$$ begin{align}
(A cap C) cap (B cap C)
&= A cap (C cap (B cap C)) tag{Associative Law} \
&= A cap ((B cap C) cap C) tag{Commutative Law} \
&= A cap (B cap (C cap C)) tag{Associative Law} \
&= A cap (B cap C) \
end{align}$$
answered Jan 21 at 8:23
BGMBGM
3,835158
$endgroup$
https://en.wikipedia.org/wiki/Algebra_of_sets
If you indeed want to play around with the commutative laws and associative laws, here we go:
$$ begin{align}
(A cap C) cap (B cap C)
&= A cap (C cap (B cap C)) tag{Associative Law} \
&= A cap ((B cap C) cap C) tag{Commutative Law} \
&= A cap (B cap (C cap C)) tag{Associative Law} \
&= A cap (B cap C) \
end{align}$$
answered Jan 21 at 8:23
BGMBGM
3,835158
answered Jan 21 at 8:23
BGMBGM
3,835158
answered Jan 21 at 8:23
BGMBGM
3,835158
answered Jan 21 at 8:23
BGMBGM
3,835158
3,835158
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
add a comment |
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
$begingroup$
Doesn't really answer my question since what you're showing isn't in MY proof, it's in the proof that someone decided to link to this question because yes, they are SIMILAR. But in order to use what you just gave me, I have to rewrite my entire proof and borrow someone elses. But since you're the only one who attempted and didn't mark it as a duplicate, I'll just give it to you. It's fine.
$endgroup$
– Sydney
Jan 21 at 19:11
1
1
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
$begingroup$
It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A cap (B cup C)) = P((A cap B) cup (A cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers.
$endgroup$
– BGM
Jan 22 at 6:34
add a comment |
