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











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












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
2






active

oldest

votes

















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













    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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














       

      draft saved


      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000797%2fis-pe-bar-f-p-bar-e-bar-f-1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      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






























                       

                      draft saved


                      draft discarded



















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000797%2fis-pe-bar-f-p-bar-e-bar-f-1%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Bundesstraße 106

                      Image
                      Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                      Read more

                      Verónica Boquete

                      Image
                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                      Read more

                      Ida-Boy-Ed-Garten

                      Image
                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                      Read more