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conditional probability of endig up in a specific set
probability of at least one person having a gem of type $n$, etc.Transitive and probabilityProbability of working machine with $3$ componentsConditioning a conditional probability to a sigma algebraConditional probability of playing a gameconditional probability about gamblingProbability of letter transmitting and receivingProbability of geometric random variables in certain orderKnockout tournament probabilityA gambler is playing a game which has a probability 0.6 to wins with $1 and probability 0.4 to lose with $1
$begingroup$
Assume we have three initial sets $mathcalA,mathcalB,mathcalC$ and 4 final sets $mathrmA,mathrmB,mathrmC,mathrmX$.
If I know that an element of $mathcalA$ can only end up with probability $a$ in $mathrmA$ and with probability $(1-a)$ in $mathrmX$. An element from $mathcalB$ can only end up with probability $b$ in $mathrmB$ and with prob. $(1-b)$ in $mathrmX$. And finally an element in $mathcalC$ can either end up with $p_1$ in $mathrmA$, with $p_2$ in $mathrmB$, with $p_3$ in $mathrmX$ or with probability $1-p_1-p_2-p_3$ in $mathrmC$.
I am a bit confused on what the conditional probability is:
$$mathbbP(i in mathcalC ; textends up in; mathrmC mid X=)$$
Is it $$frac1-(p_1+p_2+p_3)mathbbP(X=)$$ or $$frac1-(p_1+p_2)mathbbP(X=)$$ since we have that the set X is empty and therefore also no node from $mathcalC$ can be in it?
probability
asked 2 days ago
AlisatAlisat
589
$endgroup$
add a comment |
$begingroup$
Assume we have three initial sets $mathcalA,mathcalB,mathcalC$ and 4 final sets $mathrmA,mathrmB,mathrmC,mathrmX$.
If I know that an element of $mathcalA$ can only end up with probability $a$ in $mathrmA$ and with probability $(1-a)$ in $mathrmX$. An element from $mathcalB$ can only end up with probability $b$ in $mathrmB$ and with prob. $(1-b)$ in $mathrmX$. And finally an element in $mathcalC$ can either end up with $p_1$ in $mathrmA$, with $p_2$ in $mathrmB$, with $p_3$ in $mathrmX$ or with probability $1-p_1-p_2-p_3$ in $mathrmC$.
I am a bit confused on what the conditional probability is:
$$mathbbP(i in mathcalC ; textends up in; mathrmC mid X=)$$
Is it $$frac1-(p_1+p_2+p_3)mathbbP(X=)$$ or $$frac1-(p_1+p_2)mathbbP(X=)$$ since we have that the set X is empty and therefore also no node from $mathcalC$ can be in it?
probability
asked 2 days ago
AlisatAlisat
589
$endgroup$
add a comment |
$begingroup$
Assume we have three initial sets $mathcalA,mathcalB,mathcalC$ and 4 final sets $mathrmA,mathrmB,mathrmC,mathrmX$.
If I know that an element of $mathcalA$ can only end up with probability $a$ in $mathrmA$ and with probability $(1-a)$ in $mathrmX$. An element from $mathcalB$ can only end up with probability $b$ in $mathrmB$ and with prob. $(1-b)$ in $mathrmX$. And finally an element in $mathcalC$ can either end up with $p_1$ in $mathrmA$, with $p_2$ in $mathrmB$, with $p_3$ in $mathrmX$ or with probability $1-p_1-p_2-p_3$ in $mathrmC$.
I am a bit confused on what the conditional probability is:
$$mathbbP(i in mathcalC ; textends up in; mathrmC mid X=)$$
Is it $$frac1-(p_1+p_2+p_3)mathbbP(X=)$$ or $$frac1-(p_1+p_2)mathbbP(X=)$$ since we have that the set X is empty and therefore also no node from $mathcalC$ can be in it?
probability
asked 2 days ago
AlisatAlisat
589
$endgroup$
Assume we have three initial sets $mathcalA,mathcalB,mathcalC$ and 4 final sets $mathrmA,mathrmB,mathrmC,mathrmX$.
If I know that an element of $mathcalA$ can only end up with probability $a$ in $mathrmA$ and with probability $(1-a)$ in $mathrmX$. An element from $mathcalB$ can only end up with probability $b$ in $mathrmB$ and with prob. $(1-b)$ in $mathrmX$. And finally an element in $mathcalC$ can either end up with $p_1$ in $mathrmA$, with $p_2$ in $mathrmB$, with $p_3$ in $mathrmX$ or with probability $1-p_1-p_2-p_3$ in $mathrmC$.
I am a bit confused on what the conditional probability is:
$$mathbbP(i in mathcalC ; textends up in; mathrmC mid X=)$$
Is it $$frac1-(p_1+p_2+p_3)mathbbP(X=)$$ or $$frac1-(p_1+p_2)mathbbP(X=)$$ since we have that the set X is empty and therefore also no node from $mathcalC$ can be in it?
probability
probability
asked 2 days ago
AlisatAlisat
589
asked 2 days ago
AlisatAlisat
589
asked 2 days ago
AlisatAlisat
589
asked 2 days ago
AlisatAlisat
589
asked 2 days ago
AlisatAlisat
589
589
add a comment |
add a comment |
1 Answer
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$begingroup$
Your first answer is correct by a direct application of Bayes theorem.
P (i ends up in C | X empty) = P (X empty | i ends up in C) * P(i ends up in C ) / P( X empty )
What is not present in your expression is P( X empty | i ends up in C ) = 1, but that does not change the result.
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Your first answer is correct by a direct application of Bayes theorem.
P (i ends up in C | X empty) = P (X empty | i ends up in C) * P(i ends up in C ) / P( X empty )
What is not present in your expression is P( X empty | i ends up in C ) = 1, but that does not change the result.
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Your first answer is correct by a direct application of Bayes theorem.
P (i ends up in C | X empty) = P (X empty | i ends up in C) * P(i ends up in C ) / P( X empty )
What is not present in your expression is P( X empty | i ends up in C ) = 1, but that does not change the result.
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Your first answer is correct by a direct application of Bayes theorem.
P (i ends up in C | X empty) = P (X empty | i ends up in C) * P(i ends up in C ) / P( X empty )
What is not present in your expression is P( X empty | i ends up in C ) = 1, but that does not change the result.
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Your first answer is correct by a direct application of Bayes theorem.
P (i ends up in C | X empty) = P (X empty | i ends up in C) * P(i ends up in C ) / P( X empty )
What is not present in your expression is P( X empty | i ends up in C ) = 1, but that does not change the result.
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
dnqxtdnqxt
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
dnqxtdnqxt
712
answered 2 days ago
dnqxtdnqxt
712
712
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
dnqxt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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