Mixing
Proof: $P(varnothing)=0$
Clash Royale CLAN TAG#URR8PPP
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To prove(using the three probability axioms):
$$P(varnothing)=0$$
Is this method correct?
Let $A$ be an event such that $A=Omega$. Then $A^complement=varnothing$,
$$P(A)+P(A^complement)=1$$
$$P(Omega)+P(varnothing)=1tag2$$
$$P(varnothing)=0$$
Please, post any proof that is better/correct.
probability probability-theory proof-explanation
asked Jul 22 at 2:56
Zaira
132
 |Â
show 7 more comments
up vote
0
down vote
favorite
To prove(using the three probability axioms):
$$P(varnothing)=0$$
Is this method correct?
Let $A$ be an event such that $A=Omega$. Then $A^complement=varnothing$,
$$P(A)+P(A^complement)=1$$
$$P(Omega)+P(varnothing)=1tag2$$
$$P(varnothing)=0$$
Please, post any proof that is better/correct.
probability probability-theory proof-explanation
asked Jul 22 at 2:56
Zaira
132
5
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
2
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
6
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
3
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
2
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46
 |Â
show 7 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
To prove(using the three probability axioms):
$$P(varnothing)=0$$
Is this method correct?
Let $A$ be an event such that $A=Omega$. Then $A^complement=varnothing$,
$$P(A)+P(A^complement)=1$$
$$P(Omega)+P(varnothing)=1tag2$$
$$P(varnothing)=0$$
Please, post any proof that is better/correct.
probability probability-theory proof-explanation
asked Jul 22 at 2:56
Zaira
132
To prove(using the three probability axioms):
$$P(varnothing)=0$$
Is this method correct?
Let $A$ be an event such that $A=Omega$. Then $A^complement=varnothing$,
$$P(A)+P(A^complement)=1$$
$$P(Omega)+P(varnothing)=1tag2$$
$$P(varnothing)=0$$
Please, post any proof that is better/correct.
probability probability-theory proof-explanation
asked Jul 22 at 2:56
Zaira
132
edited Jul 22 at 3:33
asked Jul 22 at 2:56
Zaira
132
asked Jul 22 at 2:56
Zaira
132
asked Jul 22 at 2:56
Zaira
132
132
5
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
2
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
6
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
3
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
2
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46
 |Â
show 7 more comments
5
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
2
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
6
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
3
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
2
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46
5
5
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
2
2
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
6
6
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
3
3
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
2
2
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46
 |Â
show 7 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The method that you use is correct.
You can go for:$$1=P(S)=P(Scupvarnothing)=P(S)+P(varnothing)=1+P(varnothing)$$implying that $P(varnothing)=0$.
Here the first equality is a consequence of the axioma saying that $P(S)=1$ and the third equality is a consequence of the axioma saying that $P(Acup B)=P(A)+P(B)$ is implied by $Acap B=varnothing$.
Actually a proof that is better does not exist.
answered Jul 22 at 12:09
drhab
86.4k541118
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The method that you use is correct.
You can go for:$$1=P(S)=P(Scupvarnothing)=P(S)+P(varnothing)=1+P(varnothing)$$implying that $P(varnothing)=0$.
Here the first equality is a consequence of the axioma saying that $P(S)=1$ and the third equality is a consequence of the axioma saying that $P(Acup B)=P(A)+P(B)$ is implied by $Acap B=varnothing$.
Actually a proof that is better does not exist.
answered Jul 22 at 12:09
drhab
86.4k541118
add a comment |Â
up vote
0
down vote
accepted
The method that you use is correct.
You can go for:$$1=P(S)=P(Scupvarnothing)=P(S)+P(varnothing)=1+P(varnothing)$$implying that $P(varnothing)=0$.
Here the first equality is a consequence of the axioma saying that $P(S)=1$ and the third equality is a consequence of the axioma saying that $P(Acup B)=P(A)+P(B)$ is implied by $Acap B=varnothing$.
Actually a proof that is better does not exist.
answered Jul 22 at 12:09
drhab
86.4k541118
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The method that you use is correct.
You can go for:$$1=P(S)=P(Scupvarnothing)=P(S)+P(varnothing)=1+P(varnothing)$$implying that $P(varnothing)=0$.
Here the first equality is a consequence of the axioma saying that $P(S)=1$ and the third equality is a consequence of the axioma saying that $P(Acup B)=P(A)+P(B)$ is implied by $Acap B=varnothing$.
Actually a proof that is better does not exist.
answered Jul 22 at 12:09
drhab
86.4k541118
The method that you use is correct.
You can go for:$$1=P(S)=P(Scupvarnothing)=P(S)+P(varnothing)=1+P(varnothing)$$implying that $P(varnothing)=0$.
Here the first equality is a consequence of the axioma saying that $P(S)=1$ and the third equality is a consequence of the axioma saying that $P(Acup B)=P(A)+P(B)$ is implied by $Acap B=varnothing$.
Actually a proof that is better does not exist.
answered Jul 22 at 12:09
drhab
86.4k541118
edited Jul 22 at 12:17
answered Jul 22 at 12:09
drhab
86.4k541118
answered Jul 22 at 12:09
drhab
86.4k541118
answered Jul 22 at 12:09
drhab
86.4k541118
86.4k541118
add a comment |Â
add a comment |Â
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5
Hint: $$P(emptyset) = P(emptyset sqcup emptyset) = P(emptyset) + P(emptyset) = 2P(emptyset).$$ So...
– Dzoooks
Jul 22 at 3:01
2
@MathLover No. $sqcup$ means disjoint union. For $A neq emptyset$, we do NOT have $A = A sqcup A.$
– Dzoooks
Jul 22 at 3:04
6
@Dzoooks don't use $⊔$ to symbolise a union of disjoint sets, that's not what disjoint union means.
– Graham Kemp
Jul 22 at 3:21
3
Anyhow, Dzoooks point is that since $emptyset$ is disjoint from itself (ie $emptysetcapemptyset=emptyset$), then the additivity of probability for a union of disjoint sets means: $$mathsf P(emptysetcupemptyset)~=mathsf P(emptyset)+mathsf P(emptyset)\=2mathsf P(emptyset)$$
– Graham Kemp
Jul 22 at 3:27
2
Beware that another solution of $P(varnothing)+P(varnothing)=P(varnothing)$ is $P(varnothing)=+infty$, hence one needs an argument (for example, using the axioms of $P$) to get rid of this solution.
– Did
Jul 22 at 8:46