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How do I find (E|F')?

0

$begingroup$

Assume ' is equal to not or complement here.

Alright, you are given the following information:

p(E)= 1/3
p(F)=1/2
p(E|F)=2/5

You are asked to find p(F|E).

Bayes theorem is:

p(F|E)=p(E|F)p(F) divided by P(E|F)p(F)+p(E|F')p(F')

I know that p(E|F')= 1-p(E'|F')
How do I find p(E|F')?

I know F'=F here.

probability bayes-theorem

share|cite|improve this question

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

$endgroup$

add a comment | 

0

$begingroup$

Assume ' is equal to not or complement here.

Alright, you are given the following information:

p(E)= 1/3
p(F)=1/2
p(E|F)=2/5

You are asked to find p(F|E).

Bayes theorem is:

p(F|E)=p(E|F)p(F) divided by P(E|F)p(F)+p(E|F')p(F')

I know that p(E|F')= 1-p(E'|F')
How do I find p(E|F')?

I know F'=F here.

probability bayes-theorem

share|cite|improve this question

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

$endgroup$

add a comment | 

$begingroup$

Assume ' is equal to not or complement here.

Alright, you are given the following information:

p(E)= 1/3
p(F)=1/2
p(E|F)=2/5

You are asked to find p(F|E).

Bayes theorem is:

p(F|E)=p(E|F)p(F) divided by P(E|F)p(F)+p(E|F')p(F')

I know that p(E|F')= 1-p(E'|F')
How do I find p(E|F')?

I know F'=F here.

probability bayes-theorem

share|cite|improve this question

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

$endgroup$

p(E)= 1/3
p(F)=1/2
p(E|F)=2/5

p(F|E)=p(E|F)p(F) divided by P(E|F)p(F)+p(E|F')p(F')

share|cite|improve this question

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

share|cite|improve this question

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

asked Dec 15 '14 at 4:21

munchschairmunchschair

17610

2 Answers
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0

$begingroup$

$P(Fmid E) = dfracP(Emid F)cdot P(F)P(E) = dfracdfrac25cdot dfrac12dfrac13 = dfrac35$

share|cite|improve this answer

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

0

$begingroup$

Finding $Pr(E|F')$, as shown by the answer of ah-huh-moment, not necessary in order to find $Pr(F|E)$. That is the best possible answer to your question. But we show how to compute $Pr(E|F')$ anyway, in case we need it for something else.

Note that $Pr(Ecap F)=Pr(E|F)Pr(F)=frac15$.

But $Pr(E)=Pr(Ecap F)+Pr(Ecap F')$. Thus $Pr(Ecap F')=frac13-frac15=frac215$.

Thus $Pr(E|F')=fracPr(Ecap F')Pr(F')=frac2/151/2$.

Remark: In my experience, students do better with elementary conditional probability problems if they stick to the basic definition $Pr(A|B)=fracPr(Acap B)Pr(B)$ than if they attempt to use Bayes' Formula.

share|cite|improve this answer

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

$endgroup$

add a comment | 

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0

$begingroup$

$P(Fmid E) = dfracP(Emid F)cdot P(F)P(E) = dfracdfrac25cdot dfrac12dfrac13 = dfrac35$

share|cite|improve this answer

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

0

$begingroup$

$P(Fmid E) = dfracP(Emid F)cdot P(F)P(E) = dfracdfrac25cdot dfrac12dfrac13 = dfrac35$

share|cite|improve this answer

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

$endgroup$

add a comment | 

$begingroup$

$P(Fmid E) = dfracP(Emid F)cdot P(F)P(E) = dfracdfrac25cdot dfrac12dfrac13 = dfrac35$

share|cite|improve this answer

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

$endgroup$

share|cite|improve this answer

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

answered Dec 15 '14 at 4:26

DeepSeaDeepSea

71.4k54488

0

$begingroup$

Finding $Pr(E|F')$, as shown by the answer of ah-huh-moment, not necessary in order to find $Pr(F|E)$. That is the best possible answer to your question. But we show how to compute $Pr(E|F')$ anyway, in case we need it for something else.

Note that $Pr(Ecap F)=Pr(E|F)Pr(F)=frac15$.

But $Pr(E)=Pr(Ecap F)+Pr(Ecap F')$. Thus $Pr(Ecap F')=frac13-frac15=frac215$.

Thus $Pr(E|F')=fracPr(Ecap F')Pr(F')=frac2/151/2$.

Remark: In my experience, students do better with elementary conditional probability problems if they stick to the basic definition $Pr(A|B)=fracPr(Acap B)Pr(B)$ than if they attempt to use Bayes' Formula.

share|cite|improve this answer

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

$endgroup$

add a comment | 

0

$begingroup$

Finding $Pr(E|F')$, as shown by the answer of ah-huh-moment, not necessary in order to find $Pr(F|E)$. That is the best possible answer to your question. But we show how to compute $Pr(E|F')$ anyway, in case we need it for something else.

Note that $Pr(Ecap F)=Pr(E|F)Pr(F)=frac15$.

But $Pr(E)=Pr(Ecap F)+Pr(Ecap F')$. Thus $Pr(Ecap F')=frac13-frac15=frac215$.

Thus $Pr(E|F')=fracPr(Ecap F')Pr(F')=frac2/151/2$.

Remark: In my experience, students do better with elementary conditional probability problems if they stick to the basic definition $Pr(A|B)=fracPr(Acap B)Pr(B)$ than if they attempt to use Bayes' Formula.

share|cite|improve this answer

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

$endgroup$

add a comment | 

$begingroup$

Finding $Pr(E|F')$, as shown by the answer of ah-huh-moment, not necessary in order to find $Pr(F|E)$. That is the best possible answer to your question. But we show how to compute $Pr(E|F')$ anyway, in case we need it for something else.

Note that $Pr(Ecap F)=Pr(E|F)Pr(F)=frac15$.

But $Pr(E)=Pr(Ecap F)+Pr(Ecap F')$. Thus $Pr(Ecap F')=frac13-frac15=frac215$.

Thus $Pr(E|F')=fracPr(Ecap F')Pr(F')=frac2/151/2$.

Remark: In my experience, students do better with elementary conditional probability problems if they stick to the basic definition $Pr(A|B)=fracPr(Acap B)Pr(B)$ than if they attempt to use Bayes' Formula.

share|cite|improve this answer

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

$endgroup$

share|cite|improve this answer

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822

answered Dec 15 '14 at 5:26

André NicolasAndré Nicolas

455k36432822