Conditional probability - formula validity over different sample spaces
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Is the conditional probability formula $P(Bmid A) =frac{ P(Bcap A) }{ P(A)}$ always true? What if the sample space is non-uniform?
probability conditional-probability
edited Nov 13 at 11:00
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Is the conditional probability formula $P(Bmid A) =frac{ P(Bcap A) }{ P(A)}$ always true? What if the sample space is non-uniform?
probability conditional-probability
edited Nov 13 at 11:00
amWhy
191k27223437
asked Nov 13 at 10:34
Akash Chandwani
1033
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Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is the conditional probability formula $P(Bmid A) =frac{ P(Bcap A) }{ P(A)}$ always true? What if the sample space is non-uniform?
probability conditional-probability
edited Nov 13 at 11:00
amWhy
191k27223437
asked Nov 13 at 10:34
Akash Chandwani
1033
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is the conditional probability formula $P(Bmid A) =frac{ P(Bcap A) }{ P(A)}$ always true? What if the sample space is non-uniform?
probability conditional-probability
probability conditional-probability
edited Nov 13 at 11:00
amWhy
191k27223437
asked Nov 13 at 10:34
Akash Chandwani
1033
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 13 at 11:00
amWhy
191k27223437
asked Nov 13 at 10:34
Akash Chandwani
1033
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 13 at 11:00
amWhy
191k27223437
edited Nov 13 at 11:00
amWhy
191k27223437
edited Nov 13 at 11:00
amWhy
191k27223437
191k27223437
asked Nov 13 at 10:34
Akash Chandwani
1033
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 13 at 10:34
Akash Chandwani
1033
asked Nov 13 at 10:34
Akash Chandwani
1033
1033
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Akash Chandwani is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38
add a comment |
1
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38
1
1
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38
add a comment |
1 Answer
1
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1
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Yes, this is always true because the conditional probability of B given A is defined this way. Although, it exists only if $P(A)neq0$(that is quite obvious because if A can't occur, then it doesn't make sense to find the probability of an event given that A has already occurred).
answered Nov 13 at 11:04
Crazy for maths
4948
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, this is always true because the conditional probability of B given A is defined this way. Although, it exists only if $P(A)neq0$(that is quite obvious because if A can't occur, then it doesn't make sense to find the probability of an event given that A has already occurred).
answered Nov 13 at 11:04
Crazy for maths
4948
add a comment |
up vote
1
down vote
accepted
Yes, this is always true because the conditional probability of B given A is defined this way. Although, it exists only if $P(A)neq0$(that is quite obvious because if A can't occur, then it doesn't make sense to find the probability of an event given that A has already occurred).
answered Nov 13 at 11:04
Crazy for maths
4948
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, this is always true because the conditional probability of B given A is defined this way. Although, it exists only if $P(A)neq0$(that is quite obvious because if A can't occur, then it doesn't make sense to find the probability of an event given that A has already occurred).
answered Nov 13 at 11:04
Crazy for maths
4948
Yes, this is always true because the conditional probability of B given A is defined this way. Although, it exists only if $P(A)neq0$(that is quite obvious because if A can't occur, then it doesn't make sense to find the probability of an event given that A has already occurred).
answered Nov 13 at 11:04
Crazy for maths
4948
answered Nov 13 at 11:04
Crazy for maths
4948
answered Nov 13 at 11:04
Crazy for maths
4948
answered Nov 13 at 11:04
Crazy for maths
4948
4948
add a comment |
add a comment |
Akash Chandwani is a new contributor. Be nice, and check out our Code of Conduct.
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1
$P(B | A) =frac{P(B cap A)}{P(A)}$ is a definition, not a formula. You don't see $P(B | A)$ being defined as something else, and then that being proved equal to $frac{P(B cap A) }{P(A)}$. In particular, being a definition it very much holds in non-uniform situations, but is defined only when $P(A)$ is non-zero.
– астон вілла олоф мэллбэрг
Nov 13 at 10:38