Borel Sets and complements
Clash Royale CLAN TAG#URR8PPP
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This is a trivial question. But won't both sides of the equation be equal to zero if both events are independent? $$ P(A)P(B) - P(Acap B) = P(A^ccap B) - P(A^c)P(B)$$
And if it, how then do we show that the LHS is equivalent to the RHS? Does it suffice to say that $$P(Acap B^c) le 1-P(Acap B)$$
probability
edited Aug 23 at 2:00
Saucy O'Path
3,504324
asked Aug 23 at 1:21
Lady
356
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This is a trivial question. But won't both sides of the equation be equal to zero if both events are independent? $$ P(A)P(B) - P(Acap B) = P(A^ccap B) - P(A^c)P(B)$$
And if it, how then do we show that the LHS is equivalent to the RHS? Does it suffice to say that $$P(Acap B^c) le 1-P(Acap B)$$
probability
edited Aug 23 at 2:00
Saucy O'Path
3,504324
asked Aug 23 at 1:21
Lady
356
Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is a trivial question. But won't both sides of the equation be equal to zero if both events are independent? $$ P(A)P(B) - P(Acap B) = P(A^ccap B) - P(A^c)P(B)$$
And if it, how then do we show that the LHS is equivalent to the RHS? Does it suffice to say that $$P(Acap B^c) le 1-P(Acap B)$$
probability
edited Aug 23 at 2:00
Saucy O'Path
3,504324
asked Aug 23 at 1:21
Lady
356
This is a trivial question. But won't both sides of the equation be equal to zero if both events are independent? $$ P(A)P(B) - P(Acap B) = P(A^ccap B) - P(A^c)P(B)$$
And if it, how then do we show that the LHS is equivalent to the RHS? Does it suffice to say that $$P(Acap B^c) le 1-P(Acap B)$$
probability
edited Aug 23 at 2:00
Saucy O'Path
3,504324
asked Aug 23 at 1:21
Lady
356
edited Aug 23 at 2:00
Saucy O'Path
3,504324
edited Aug 23 at 2:00
Saucy O'Path
3,504324
edited Aug 23 at 2:00
Saucy O'Path
3,504324
3,504324
asked Aug 23 at 1:21
Lady
356
asked Aug 23 at 1:21
Lady
356
asked Aug 23 at 1:21
Lady
356
356
Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01
add a comment |Â
Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01
Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01
Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01
add a comment |Â
2 Answers
2
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oldest
votes
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1
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Yes, they will.
$P(Bsetminus A)-P(Xsetminus A)P(B)=P(B)-P(Bcap A)-(1-P(A))P(B)=\=P(A)P(B)-P(Bcap A)$
No, it doesn't.
answered Aug 23 at 1:55
Saucy O'Path
3,504324
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
add a comment |Â
up vote
0
down vote
The equation is equivalent to $$P(A^c)P(B)+P(A)P(B)=P(Acap B)+P(A^ccap B).$$ The LHS of this is $(P(A^c)+P(A))P(B)=(1)P(B)=P(B).$
I will leave it to you to confirm that the RHS, $P(Acap B)+P(A^ccap B)$ is also equal to $P(B).$
answered Aug 23 at 5:58
DanielWainfleet
32k31644
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes, they will.
$P(Bsetminus A)-P(Xsetminus A)P(B)=P(B)-P(Bcap A)-(1-P(A))P(B)=\=P(A)P(B)-P(Bcap A)$
No, it doesn't.
answered Aug 23 at 1:55
Saucy O'Path
3,504324
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
add a comment |Â
up vote
1
down vote
Yes, they will.
$P(Bsetminus A)-P(Xsetminus A)P(B)=P(B)-P(Bcap A)-(1-P(A))P(B)=\=P(A)P(B)-P(Bcap A)$
No, it doesn't.
answered Aug 23 at 1:55
Saucy O'Path
3,504324
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, they will.
$P(Bsetminus A)-P(Xsetminus A)P(B)=P(B)-P(Bcap A)-(1-P(A))P(B)=\=P(A)P(B)-P(Bcap A)$
No, it doesn't.
answered Aug 23 at 1:55
Saucy O'Path
3,504324
Yes, they will.
$P(Bsetminus A)-P(Xsetminus A)P(B)=P(B)-P(Bcap A)-(1-P(A))P(B)=\=P(A)P(B)-P(Bcap A)$
No, it doesn't.
answered Aug 23 at 1:55
Saucy O'Path
3,504324
answered Aug 23 at 1:55
Saucy O'Path
3,504324
answered Aug 23 at 1:55
Saucy O'Path
3,504324
answered Aug 23 at 1:55
Saucy O'Path
3,504324
3,504324
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
add a comment |Â
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
how then do we prove $$P(Acap B^c) le 1-P(Acap B)$$
– Lady
Aug 23 at 5:02
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
Prove that $P(A^ccap B)+P(Acap B)=P(B). $
– DanielWainfleet
Aug 23 at 6:04
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
@Lady Ah, so it was "prove the inequality from the identity" rather than "prove the identity from the inequality". Truth to be told, that inequality is an immediate consequence of some tools I've used in point (2). Since they are very basilar, the exercise for you is to spot which ones and why.
– Saucy O'Path
Aug 23 at 7:54
add a comment |Â
up vote
0
down vote
The equation is equivalent to $$P(A^c)P(B)+P(A)P(B)=P(Acap B)+P(A^ccap B).$$ The LHS of this is $(P(A^c)+P(A))P(B)=(1)P(B)=P(B).$
I will leave it to you to confirm that the RHS, $P(Acap B)+P(A^ccap B)$ is also equal to $P(B).$
answered Aug 23 at 5:58
DanielWainfleet
32k31644
add a comment |Â
up vote
0
down vote
The equation is equivalent to $$P(A^c)P(B)+P(A)P(B)=P(Acap B)+P(A^ccap B).$$ The LHS of this is $(P(A^c)+P(A))P(B)=(1)P(B)=P(B).$
I will leave it to you to confirm that the RHS, $P(Acap B)+P(A^ccap B)$ is also equal to $P(B).$
answered Aug 23 at 5:58
DanielWainfleet
32k31644
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The equation is equivalent to $$P(A^c)P(B)+P(A)P(B)=P(Acap B)+P(A^ccap B).$$ The LHS of this is $(P(A^c)+P(A))P(B)=(1)P(B)=P(B).$
I will leave it to you to confirm that the RHS, $P(Acap B)+P(A^ccap B)$ is also equal to $P(B).$
answered Aug 23 at 5:58
DanielWainfleet
32k31644
The equation is equivalent to $$P(A^c)P(B)+P(A)P(B)=P(Acap B)+P(A^ccap B).$$ The LHS of this is $(P(A^c)+P(A))P(B)=(1)P(B)=P(B).$
I will leave it to you to confirm that the RHS, $P(Acap B)+P(A^ccap B)$ is also equal to $P(B).$
answered Aug 23 at 5:58
DanielWainfleet
32k31644
answered Aug 23 at 5:58
DanielWainfleet
32k31644
answered Aug 23 at 5:58
DanielWainfleet
32k31644
answered Aug 23 at 5:58
DanielWainfleet
32k31644
32k31644
add a comment |Â
add a comment |Â
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Your are right that if A,B are independent then both sides are 0. But the equation holds for all A,B.
– DanielWainfleet
Aug 23 at 6:01