Mixing
Proof of: If $P(A) = P(B) = 1$ then $P(A cap B) = 1$.
$begingroup$
Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(Acap B)= P(A) cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1cdot 1=1$ so $P(Acap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks
probability proof-verification
edited Jan 23 at 19:07
MathOverview
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
$endgroup$
add a comment |
$begingroup$
Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(Acap B)= P(A) cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1cdot 1=1$ so $P(Acap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks
probability proof-verification
edited Jan 23 at 19:07
MathOverview
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
$endgroup$
$begingroup$
The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
$endgroup$
– vadim123
Jan 23 at 15:32
$begingroup$
The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
$endgroup$
– AnoUser1
Jan 23 at 15:35
$begingroup$
@AnoUser1, you can ask that as a separate question.
$endgroup$
– vadim123
Jan 23 at 15:38
$begingroup$
Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 23 at 15:38
add a comment |
$begingroup$
Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(Acap B)= P(A) cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1cdot 1=1$ so $P(Acap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks
probability proof-verification
edited Jan 23 at 19:07
MathOverview
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
$endgroup$
Ok so I know this is obviously true but not sure if my method is right to proving it. Here's my go.
$P(Acap B)= P(A) cdot P(B)$ by definition.
$P(A)=P(B)=1$
Therefore $1cdot 1=1$ so $P(Acap B)=1$
I realise this is a really simply question but this answer seems too easy.
Thanks
probability proof-verification
probability proof-verification
edited Jan 23 at 19:07
MathOverview
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
edited Jan 23 at 19:07
MathOverview
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
edited Jan 23 at 19:07
MathOverview
8,94543164
edited Jan 23 at 19:07
MathOverview
8,94543164
edited Jan 23 at 19:07
MathOverview
8,94543164
8,94543164
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
asked Jan 23 at 15:29
AnoUser1AnoUser1
836
836
$begingroup$
The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
$endgroup$
– vadim123
Jan 23 at 15:32
$begingroup$
The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
$endgroup$
– AnoUser1
Jan 23 at 15:35
$begingroup$
@AnoUser1, you can ask that as a separate question.
$endgroup$
– vadim123
Jan 23 at 15:38
$begingroup$
Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 23 at 15:38
add a comment |
$begingroup$
The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
$endgroup$
– vadim123
Jan 23 at 15:32
$begingroup$
The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
$endgroup$
– AnoUser1
Jan 23 at 15:35
$begingroup$
@AnoUser1, you can ask that as a separate question.
$endgroup$
– vadim123
Jan 23 at 15:38
$begingroup$
Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 23 at 15:38
$begingroup$
The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
$endgroup$
– vadim123
Jan 23 at 15:32
$begingroup$
The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
$endgroup$
– vadim123
Jan 23 at 15:32
$begingroup$
The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
$endgroup$
– AnoUser1
Jan 23 at 15:35
$begingroup$
The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
$endgroup$
– AnoUser1
Jan 23 at 15:35
$begingroup$
@AnoUser1, you can ask that as a separate question.
$endgroup$
– vadim123
Jan 23 at 15:38
$begingroup$
@AnoUser1, you can ask that as a separate question.
$endgroup$
– vadim123
Jan 23 at 15:38
$begingroup$
Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 23 at 15:38
$begingroup$
Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 23 at 15:38
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Well, you cannot say $P(Acap B)= P(A) cdot P(B)$ because you do not know if $A$ and $B$ are independent.
Anyway, you could say $$P(Acap B) = 1-P((Acap B)^c)=1-P(A^c cup B^c) geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
answered Jan 23 at 15:35
franzkfranzk
1435
$endgroup$
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
add a comment |
$begingroup$
We begin with $$P(Acup B)=P(A)+P(B)-P(Acap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as
$$P(Acap B)=2-P(Acup B)$$
While we might not know $P(Acup B)$, we know that $P(Acup B)le 1$ (true for all probabilities), and hence $$P(Acap B)ge 2-1=1$$
Since also $P(Acap B)le 1$ (true for all probabilities), we can be sure that $P(Acap B)=1$.
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
$begingroup$
Note that $Asubset Acup B$ and $Bsubset Acup B$ implies $P(Acup B)geq P(A)$ and $P(Acup B)geq P(B)$. Then $P(Acup B)=1$. By Inclusion–exclusion principle
$$
P(Acup B)= P(A)+P(B)-P(Acap B).
$$
So
$$
1=1+1-P(Acap B).
$$
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
$endgroup$
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
add a comment |
$begingroup$
$P(A|B)=frac{P(A bigcap B)}{P(B)}$, thus $P(A bigcap B)=P(A|B)P(B)$.
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A bigcap B)=1$.
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, you cannot say $P(Acap B)= P(A) cdot P(B)$ because you do not know if $A$ and $B$ are independent.
Anyway, you could say $$P(Acap B) = 1-P((Acap B)^c)=1-P(A^c cup B^c) geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
answered Jan 23 at 15:35
franzkfranzk
1435
$endgroup$
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
add a comment |
$begingroup$
Well, you cannot say $P(Acap B)= P(A) cdot P(B)$ because you do not know if $A$ and $B$ are independent.
Anyway, you could say $$P(Acap B) = 1-P((Acap B)^c)=1-P(A^c cup B^c) geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
answered Jan 23 at 15:35
franzkfranzk
1435
$endgroup$
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
add a comment |
$begingroup$
Well, you cannot say $P(Acap B)= P(A) cdot P(B)$ because you do not know if $A$ and $B$ are independent.
Anyway, you could say $$P(Acap B) = 1-P((Acap B)^c)=1-P(A^c cup B^c) geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
answered Jan 23 at 15:35
franzkfranzk
1435
$endgroup$
Well, you cannot say $P(Acap B)= P(A) cdot P(B)$ because you do not know if $A$ and $B$ are independent.
Anyway, you could say $$P(Acap B) = 1-P((Acap B)^c)=1-P(A^c cup B^c) geq 1-(P(A^c)+ P(B^c))= 1-(0+0)=1$$
answered Jan 23 at 15:35
franzkfranzk
1435
answered Jan 23 at 15:35
franzkfranzk
1435
answered Jan 23 at 15:35
franzkfranzk
1435
answered Jan 23 at 15:35
franzkfranzk
1435
1435
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
add a comment |
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
3
3
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
$begingroup$
(which of course proves that actually, $A$ and $B$ were in fact independent :) )
$endgroup$
– Calvin Khor
Jan 23 at 19:14
add a comment |
$begingroup$
We begin with $$P(Acup B)=P(A)+P(B)-P(Acap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as
$$P(Acap B)=2-P(Acup B)$$
While we might not know $P(Acup B)$, we know that $P(Acup B)le 1$ (true for all probabilities), and hence $$P(Acap B)ge 2-1=1$$
Since also $P(Acap B)le 1$ (true for all probabilities), we can be sure that $P(Acap B)=1$.
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
$begingroup$
We begin with $$P(Acup B)=P(A)+P(B)-P(Acap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as
$$P(Acap B)=2-P(Acup B)$$
While we might not know $P(Acup B)$, we know that $P(Acup B)le 1$ (true for all probabilities), and hence $$P(Acap B)ge 2-1=1$$
Since also $P(Acap B)le 1$ (true for all probabilities), we can be sure that $P(Acap B)=1$.
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
$begingroup$
We begin with $$P(Acup B)=P(A)+P(B)-P(Acap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as
$$P(Acap B)=2-P(Acup B)$$
While we might not know $P(Acup B)$, we know that $P(Acup B)le 1$ (true for all probabilities), and hence $$P(Acap B)ge 2-1=1$$
Since also $P(Acap B)le 1$ (true for all probabilities), we can be sure that $P(Acap B)=1$.
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
$endgroup$
We begin with $$P(Acup B)=P(A)+P(B)-P(Acap B)$$
Since $P(A)=P(B)=1$, we can rearrange this as
$$P(Acap B)=2-P(Acup B)$$
While we might not know $P(Acup B)$, we know that $P(Acup B)le 1$ (true for all probabilities), and hence $$P(Acap B)ge 2-1=1$$
Since also $P(Acap B)le 1$ (true for all probabilities), we can be sure that $P(Acap B)=1$.
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
answered Jan 23 at 15:37
vadim123vadim123
76.4k897191
76.4k897191
add a comment |
add a comment |
$begingroup$
Note that $Asubset Acup B$ and $Bsubset Acup B$ implies $P(Acup B)geq P(A)$ and $P(Acup B)geq P(B)$. Then $P(Acup B)=1$. By Inclusion–exclusion principle
$$
P(Acup B)= P(A)+P(B)-P(Acap B).
$$
So
$$
1=1+1-P(Acap B).
$$
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
$endgroup$
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
add a comment |
$begingroup$
Note that $Asubset Acup B$ and $Bsubset Acup B$ implies $P(Acup B)geq P(A)$ and $P(Acup B)geq P(B)$. Then $P(Acup B)=1$. By Inclusion–exclusion principle
$$
P(Acup B)= P(A)+P(B)-P(Acap B).
$$
So
$$
1=1+1-P(Acap B).
$$
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
$endgroup$
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
add a comment |
$begingroup$
Note that $Asubset Acup B$ and $Bsubset Acup B$ implies $P(Acup B)geq P(A)$ and $P(Acup B)geq P(B)$. Then $P(Acup B)=1$. By Inclusion–exclusion principle
$$
P(Acup B)= P(A)+P(B)-P(Acap B).
$$
So
$$
1=1+1-P(Acap B).
$$
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
$endgroup$
Note that $Asubset Acup B$ and $Bsubset Acup B$ implies $P(Acup B)geq P(A)$ and $P(Acup B)geq P(B)$. Then $P(Acup B)=1$. By Inclusion–exclusion principle
$$
P(Acup B)= P(A)+P(B)-P(Acap B).
$$
So
$$
1=1+1-P(Acap B).
$$
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
edited Jan 23 at 17:32
Paul Sinclair
20.4k21443
20.4k21443
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
answered Jan 23 at 15:41
MathOverviewMathOverview
8,94543164
8,94543164
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
add a comment |
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
I can't make an edit that small but you're missing a B in $Bsubset Acup B$ (first line)
$endgroup$
– Droplet
Jan 23 at 17:27
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
$begingroup$
@Droplet - I can, Thanks for pointing it out.
$endgroup$
– Paul Sinclair
Jan 23 at 17:32
add a comment |
$begingroup$
$P(A|B)=frac{P(A bigcap B)}{P(B)}$, thus $P(A bigcap B)=P(A|B)P(B)$.
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A bigcap B)=1$.
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
$P(A|B)=frac{P(A bigcap B)}{P(B)}$, thus $P(A bigcap B)=P(A|B)P(B)$.
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A bigcap B)=1$.
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
$endgroup$
add a comment |
$begingroup$
$P(A|B)=frac{P(A bigcap B)}{P(B)}$, thus $P(A bigcap B)=P(A|B)P(B)$.
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A bigcap B)=1$.
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
$endgroup$
$P(A|B)=frac{P(A bigcap B)}{P(B)}$, thus $P(A bigcap B)=P(A|B)P(B)$.
If $P(A)=1$, then that means that $A$ always happens, so it will still always happen when $B$ happens, so $P(A|B)=1$. Since $P(B)=1$, $P(A bigcap B)=1$.
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
answered Jan 29 at 20:43
AcccumulationAcccumulation
7,1252619
7,1252619
add a comment |
add a comment |
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The statement $P(Acap B)=P(A)P(B)$ is true only for independent events $A,B$. We don't know that's true.
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– vadim123
Jan 23 at 15:32
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The question also says ''Let Ai , i = 1, 2, 3, . . . be a collection of events in the sample space Ω. Prove the following statements:'' at the start. So given they're constantly P(Ai)=1, we know they're independent?
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– AnoUser1
Jan 23 at 15:35
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@AnoUser1, you can ask that as a separate question.
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– vadim123
Jan 23 at 15:38
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Similarly to the answer below : $1 ge P(A cup B)=P(A)+P(B) - P(A cap B)=2 - P(A cap B) ge 1$. From this : $P(A cup B)=1$ and then $P(A cap B)=1$.
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– Mauro ALLEGRANZA
Jan 23 at 15:38