Is $P(E/bar F) + P(bar E/ bar F) = 1$?











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1
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Is $P(E/bar F) + P(bar E/ bar F) = 1$?



I tried using venn diagram and here's what I get:



enter image description here



The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$



Clearly it is not $1$ but my book says it is. Where have I misunderstood?






probability







share|cite|improve this question






















  • You probably confuse the relative complement and the conditional probability.
    – Hanul Jeon
    2 days ago










  • You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
    – астон вілла олоф мэллбэрг
    2 days ago












  • Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
    – Lord Shark the Unknown
    2 days ago










  • Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
    – Abcd
    2 days ago










  • @Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
    – trancelocation
    2 days ago















up vote
1
down vote

favorite












Is $P(E/bar F) + P(bar E/ bar F) = 1$?



I tried using venn diagram and here's what I get:



enter image description here



The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$



Clearly it is not $1$ but my book says it is. Where have I misunderstood?






probability







share|cite|improve this question






















  • You probably confuse the relative complement and the conditional probability.
    – Hanul Jeon
    2 days ago










  • You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
    – астон вілла олоф мэллбэрг
    2 days ago












  • Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
    – Lord Shark the Unknown
    2 days ago










  • Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
    – Abcd
    2 days ago










  • @Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
    – trancelocation
    2 days ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Is $P(E/bar F) + P(bar E/ bar F) = 1$?



I tried using venn diagram and here's what I get:



enter image description here



The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$



Clearly it is not $1$ but my book says it is. Where have I misunderstood?






probability







share|cite|improve this question













Is $P(E/bar F) + P(bar E/ bar F) = 1$?



I tried using venn diagram and here's what I get:



enter image description here



The red region shows $P(E/ bar F)$ and the blue region shows $P(bar E/ bar F)$



Clearly it is not $1$ but my book says it is. Where have I misunderstood?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Abcd

2,87311130




2,87311130












  • You probably confuse the relative complement and the conditional probability.
    – Hanul Jeon
    2 days ago










  • You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
    – астон вілла олоф мэллбэрг
    2 days ago












  • Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
    – Lord Shark the Unknown
    2 days ago










  • Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
    – Abcd
    2 days ago










  • @Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
    – trancelocation
    2 days ago


















  • You probably confuse the relative complement and the conditional probability.
    – Hanul Jeon
    2 days ago










  • You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
    – астон вілла олоф мэллбэрг
    2 days ago












  • Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
    – Lord Shark the Unknown
    2 days ago










  • Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
    – Abcd
    2 days ago










  • @Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
    – trancelocation
    2 days ago
















You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
2 days ago




You probably confuse the relative complement and the conditional probability.
– Hanul Jeon
2 days ago












You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
2 days ago






You are confusing $/$ , that is conditional probability, with $cap$, the intersection. Conditional probability $P(E / F)$ requires that you take the ratio $frac{P(E cap F)}{P(F)}$, so it is not the probability of some event in the space, but rather a ratio of two probabilities, which can't be represented on the Venn diagram.
– астон вілла олоф мэллбэрг
2 days ago














Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
2 days ago




Is $P(E/F)$ a conditional probability. If so, then recall the definition of conditional probability.
– Lord Shark the Unknown
2 days ago












Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
2 days ago




Is it not possible to sketch conditional probability on venn diagram @LordSharktheUnknown
– Abcd
2 days ago












@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
2 days ago




@Abcd You can sketch it. Just cut out the circle for $F$. Now, this is your new sample space.
– trancelocation
2 days ago










2 Answers
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Note that





  • $(E cap bar F) cup (bar E cap bar F) = bar F$ and

  • $(E cap bar F) cap (bar E cap bar F) = emptyset$






share|cite|improve this answer




























    up vote
    0
    down vote













    Since $E$ and $bar{E}$ are disjoint events we have:



    $$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$



    The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
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      up vote
      0
      down vote













      Note that





      • $(E cap bar F) cup (bar E cap bar F) = bar F$ and

      • $(E cap bar F) cap (bar E cap bar F) = emptyset$






      share|cite|improve this answer

























        up vote
        0
        down vote













        Note that





        • $(E cap bar F) cup (bar E cap bar F) = bar F$ and

        • $(E cap bar F) cap (bar E cap bar F) = emptyset$






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Note that





          • $(E cap bar F) cup (bar E cap bar F) = bar F$ and

          • $(E cap bar F) cap (bar E cap bar F) = emptyset$






          share|cite|improve this answer












          Note that





          • $(E cap bar F) cup (bar E cap bar F) = bar F$ and

          • $(E cap bar F) cap (bar E cap bar F) = emptyset$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          trancelocation

          7,8461519




          7,8461519






















              up vote
              0
              down vote













              Since $E$ and $bar{E}$ are disjoint events we have:



              $$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$



              The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).






              share|cite|improve this answer

























                up vote
                0
                down vote













                Since $E$ and $bar{E}$ are disjoint events we have:



                $$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$



                The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Since $E$ and $bar{E}$ are disjoint events we have:



                  $$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$



                  The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).






                  share|cite|improve this answer












                  Since $E$ and $bar{E}$ are disjoint events we have:



                  $$mathbb{P}(E|bar{F}) + mathbb{P}(bar{E}|bar{F}) = mathbb{P}(E cup bar{E}|bar{F}) = mathbb{P}(Omega|bar{F}) = 1.$$



                  The first step follows from the additivity axiom of probability and the last follows from the norming axiom. Your diagram does not accurately show the conditional probabilities: the first is the relative size of the red area in the union of both the red and blue areas, and the second is the relative size of the blue area in the union of both the red and blue areas, which sum to the whole area (i.e., one).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Ben

                  87311




                  87311






























                       

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