Finding how 3 hunter can share their prey












0












$begingroup$


Three hunters hit the moving target with probabilities $0,6, 0,4$ and $0,2$. When they saw a deer they shot in one time. After it they saw that only one bullet reached the target. So how they can shere a prey?



I think first I need to find $mathbb{P}(X)$ where $X$ is event that deer was shot. $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C)) $ am I right? When I will find it I will make a proportion and will find how they can share a prey.



So if I am right, I have problem. I think I am doing something wrong.
$mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C))=
mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C})) +mathbb{P} (overline{A}cap overline{B} cap C) - mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))=
mathbb{P}(Acap overline{B}cap overline{C}) + mathbb{P}((overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))-mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+mathbb{P} (overline{A}cap overline{B} cap C) -
mathbb{P} (Acap overline{B}cap overline{C}) - mathbb{P}( (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+ mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))= ?$ Where is my mistake and how I can continue it? Maybe there is easier way?






probability probability-theory random-variables







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Three hunters hit the moving target with probabilities $0,6, 0,4$ and $0,2$. When they saw a deer they shot in one time. After it they saw that only one bullet reached the target. So how they can shere a prey?



    I think first I need to find $mathbb{P}(X)$ where $X$ is event that deer was shot. $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C)) $ am I right? When I will find it I will make a proportion and will find how they can share a prey.



    So if I am right, I have problem. I think I am doing something wrong.
    $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C))=
    mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C})) +mathbb{P} (overline{A}cap overline{B} cap C) - mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))=
    mathbb{P}(Acap overline{B}cap overline{C}) + mathbb{P}((overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))-mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+mathbb{P} (overline{A}cap overline{B} cap C) -
    mathbb{P} (Acap overline{B}cap overline{C}) - mathbb{P}( (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+ mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))= ?$     Where is my mistake and how I can continue it? Maybe there is easier way?






    probability probability-theory random-variables







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Three hunters hit the moving target with probabilities $0,6, 0,4$ and $0,2$. When they saw a deer they shot in one time. After it they saw that only one bullet reached the target. So how they can shere a prey?



      I think first I need to find $mathbb{P}(X)$ where $X$ is event that deer was shot. $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C)) $ am I right? When I will find it I will make a proportion and will find how they can share a prey.



      So if I am right, I have problem. I think I am doing something wrong.
      $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C))=
      mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C})) +mathbb{P} (overline{A}cap overline{B} cap C) - mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))=
      mathbb{P}(Acap overline{B}cap overline{C}) + mathbb{P}((overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))-mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+mathbb{P} (overline{A}cap overline{B} cap C) -
      mathbb{P} (Acap overline{B}cap overline{C}) - mathbb{P}( (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+ mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))= ?$       Where is my mistake and how I can continue it? Maybe there is easier way?






      probability probability-theory random-variables







      share|cite|improve this question









      $endgroup$




      Three hunters hit the moving target with probabilities $0,6, 0,4$ and $0,2$. When they saw a deer they shot in one time. After it they saw that only one bullet reached the target. So how they can shere a prey?



      I think first I need to find $mathbb{P}(X)$ where $X$ is event that deer was shot. $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C)) $ am I right? When I will find it I will make a proportion and will find how they can share a prey.



      So if I am right, I have problem. I think I am doing something wrong.
      $mathbb{P}(X)=mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cup (overline{A}cap overline{B} cap C))=
      mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C})) +mathbb{P} (overline{A}cap overline{B} cap C) - mathbb{P}( (Acap overline{B}cap overline{C}) cup (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))=
      mathbb{P}(Acap overline{B}cap overline{C}) + mathbb{P}((overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))-mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+mathbb{P} (overline{A}cap overline{B} cap C) -
      mathbb{P} (Acap overline{B}cap overline{C}) - mathbb{P}( (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))+ mathbb{P}( (Acap overline{B}cap overline{C}) cap (overline{A}cap B cap overline{C}) cap (overline{A}cap overline{B} cap C))= ?$       Where is my mistake and how I can continue it? Maybe there is easier way?






      probability probability-theory random-variables




      probability probability-theory random-variables






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 11 '18 at 4:49









      AtstovasAtstovas

      1139




      1139






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          $X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6cdot (1-0.4) cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
            $endgroup$
            – Atstovas
            Dec 11 '18 at 5:45












          • $begingroup$
            Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
            $endgroup$
            – Ross Millikan
            Dec 11 '18 at 5:54














          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',
          autoActivateHeartbeat: false,
          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%2f3034898%2ffinding-how-3-hunter-can-share-their-prey%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          $X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6cdot (1-0.4) cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
            $endgroup$
            – Atstovas
            Dec 11 '18 at 5:45












          • $begingroup$
            Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
            $endgroup$
            – Ross Millikan
            Dec 11 '18 at 5:54


















          1












          $begingroup$

          $X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6cdot (1-0.4) cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
            $endgroup$
            – Atstovas
            Dec 11 '18 at 5:45












          • $begingroup$
            Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
            $endgroup$
            – Ross Millikan
            Dec 11 '18 at 5:54
















          1












          1








          1





          $begingroup$

          $X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6cdot (1-0.4) cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.






          share|cite|improve this answer









          $endgroup$



          $X$ should be the event the deer was shot exactly once, because that is the information you have. You are trying to assess $P(A|X)$, the chance that $A$ hit the deer given that exactly one person hit the deer, and the same for the other two. The probability that $A$ hit and the others did not is $0.6cdot (1-0.4) cdot (1-0.2)=0.288$. Compute the corresponding probability for the other two and scale up to make the sum be $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 4:57









          Ross MillikanRoss Millikan

          300k24200375




          300k24200375












          • $begingroup$
            So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
            $endgroup$
            – Atstovas
            Dec 11 '18 at 5:45












          • $begingroup$
            Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
            $endgroup$
            – Ross Millikan
            Dec 11 '18 at 5:54




















          • $begingroup$
            So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
            $endgroup$
            – Atstovas
            Dec 11 '18 at 5:45












          • $begingroup$
            Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
            $endgroup$
            – Ross Millikan
            Dec 11 '18 at 5:54


















          $begingroup$
          So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
          $endgroup$
          – Atstovas
          Dec 11 '18 at 5:45






          $begingroup$
          So $mathbb{P}(A|X)=0.288$ $mathbb{P}(B|X)=0.192$ $mathbb{P}(C|X)=0.048$ than $0.288+0.192+0.048=0,528$ Whats next? For first hunter it will be $ 0.288/0.528=0.(54)$ for second $0.192/0.528=0.(36)$ and for third hunter $0.048/0.528=0.(09)$ right?
          $endgroup$
          – Atstovas
          Dec 11 '18 at 5:45














          $begingroup$
          Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
          $endgroup$
          – Ross Millikan
          Dec 11 '18 at 5:54






          $begingroup$
          Your result is correct but your notation is wrong. Based on your calculation $P(A|X)=frac {54}{99}$ which is what you mean by $0.(54)$
          $endgroup$
          – Ross Millikan
          Dec 11 '18 at 5:54




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034898%2ffinding-how-3-hunter-can-share-their-prey%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

          How to send String Array data to Server using php in android

          Image
          0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
          Read more

          Is anime1.com a legal site for watching anime?

          Image
          up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
          Read more