Experiment with negative binomial distribution












0












$begingroup$


Suppose I toss a coin with $0.25$ chance of $H$. I toss it until I get $k+1$ times heads. What is the probability that I will have to toss the coin $4k$ times?



The last throw must be an $H$ so I have $4k-1$ options to choose $k-1$ throws where I get $H$. So $$ { 4k-1 choose k-1 }cdot 0.75^{3k-1} cdot 0.25 ^{k}$$



But I'm wrong about something here as the answer is $ { 4k-1 choose k } cdot 0.75^{3k-1} cdot 0.25^{k+1} $



I know I can just plug the variables into negative binomial formula but still I don't understand what am I missing here? what I need $ 0.25^{k+1} $ successes? and why I need to chose $ k $ places instead of $ k-1 $ places as I know the last throw must be $H$ ??



Thanks!






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
    $endgroup$
    – drhab
    Dec 6 '18 at 15:11


















0












$begingroup$


Suppose I toss a coin with $0.25$ chance of $H$. I toss it until I get $k+1$ times heads. What is the probability that I will have to toss the coin $4k$ times?



The last throw must be an $H$ so I have $4k-1$ options to choose $k-1$ throws where I get $H$. So $$ { 4k-1 choose k-1 }cdot 0.75^{3k-1} cdot 0.25 ^{k}$$



But I'm wrong about something here as the answer is $ { 4k-1 choose k } cdot 0.75^{3k-1} cdot 0.25^{k+1} $



I know I can just plug the variables into negative binomial formula but still I don't understand what am I missing here? what I need $ 0.25^{k+1} $ successes? and why I need to chose $ k $ places instead of $ k-1 $ places as I know the last throw must be $H$ ??



Thanks!






probability







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
    $endgroup$
    – drhab
    Dec 6 '18 at 15:11
















0












0








0


1



$begingroup$


Suppose I toss a coin with $0.25$ chance of $H$. I toss it until I get $k+1$ times heads. What is the probability that I will have to toss the coin $4k$ times?



The last throw must be an $H$ so I have $4k-1$ options to choose $k-1$ throws where I get $H$. So $$ { 4k-1 choose k-1 }cdot 0.75^{3k-1} cdot 0.25 ^{k}$$



But I'm wrong about something here as the answer is $ { 4k-1 choose k } cdot 0.75^{3k-1} cdot 0.25^{k+1} $



I know I can just plug the variables into negative binomial formula but still I don't understand what am I missing here? what I need $ 0.25^{k+1} $ successes? and why I need to chose $ k $ places instead of $ k-1 $ places as I know the last throw must be $H$ ??



Thanks!






probability







share|cite|improve this question











$endgroup$




Suppose I toss a coin with $0.25$ chance of $H$. I toss it until I get $k+1$ times heads. What is the probability that I will have to toss the coin $4k$ times?



The last throw must be an $H$ so I have $4k-1$ options to choose $k-1$ throws where I get $H$. So $$ { 4k-1 choose k-1 }cdot 0.75^{3k-1} cdot 0.25 ^{k}$$



But I'm wrong about something here as the answer is $ { 4k-1 choose k } cdot 0.75^{3k-1} cdot 0.25^{k+1} $



I know I can just plug the variables into negative binomial formula but still I don't understand what am I missing here? what I need $ 0.25^{k+1} $ successes? and why I need to chose $ k $ places instead of $ k-1 $ places as I know the last throw must be $H$ ??



Thanks!






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 15:29









Math Girl

631318




631318










asked Dec 6 '18 at 14:58









bm1125bm1125

64016




64016








  • 1




    $begingroup$
    You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
    $endgroup$
    – drhab
    Dec 6 '18 at 15:11
















  • 1




    $begingroup$
    You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
    $endgroup$
    – drhab
    Dec 6 '18 at 15:11










1




1




$begingroup$
You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
$endgroup$
– drhab
Dec 6 '18 at 15:11






$begingroup$
You are almost correct. Note that in total $k+1$ heads (successes) are needed (not $k$): exactly $k$ in the first $4k-1$ throws and $1$ at the $4k$-th throw.
$endgroup$
– drhab
Dec 6 '18 at 15:11












2 Answers
2






active

oldest

votes


















1












$begingroup$

If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is



$$ P(textrm{there are } k textrm{ heads out of the first } 4k-1 textrm{ throws})= p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k} $$



Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So



$$ P(textrm{it takes 4k throws to see } k+1 textrm{ heads}) = 0.25p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k+1}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.






    share|cite|improve this answer









    $endgroup$













      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%2f3028598%2fexperiment-with-negative-binomial-distribution%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









      1












      $begingroup$

      If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is



      $$ P(textrm{there are } k textrm{ heads out of the first } 4k-1 textrm{ throws})= p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k} $$



      Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So



      $$ P(textrm{it takes 4k throws to see } k+1 textrm{ heads}) = 0.25p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k+1}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is



        $$ P(textrm{there are } k textrm{ heads out of the first } 4k-1 textrm{ throws})= p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k} $$



        Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So



        $$ P(textrm{it takes 4k throws to see } k+1 textrm{ heads}) = 0.25p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k+1}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is



          $$ P(textrm{there are } k textrm{ heads out of the first } 4k-1 textrm{ throws})= p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k} $$



          Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So



          $$ P(textrm{it takes 4k throws to see } k+1 textrm{ heads}) = 0.25p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k+1}$$






          share|cite|improve this answer









          $endgroup$



          If you need $k+1$ heads overall then you must succeed $k$ times from the first $4k-1$ throws - if you throw $k-1$ heads in the first $4k-1$ and then another head then that is only $k$ heads overall. The probability that you succeed $k$ times in the first $4k-1$ throws is



          $$ P(textrm{there are } k textrm{ heads out of the first } 4k-1 textrm{ throws})= p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k} $$



          Now, the probability that it takes you $4k$ throws to succeed $k+1$ times is the probability that you throw $k$ heads in the first $4k-1$ throws times the probability the last throw is a head (which is 0.25). So



          $$ P(textrm{it takes 4k throws to see } k+1 textrm{ heads}) = 0.25p = binom{4k-1}{k}cdot 0.75^{3k-1} cdot 0.25^{k+1}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 15:07









          ODFODF

          1,461510




          1,461510























              1












              $begingroup$

              Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.






                  share|cite|improve this answer









                  $endgroup$



                  Since your last throw results in a head and you need a total of $k+1$ heads, you have $4k-1$ throws to get the rest of $k$ heads.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 15:06









                  Thomas ShelbyThomas Shelby

                  2,585421




                  2,585421






























                      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%2f3028598%2fexperiment-with-negative-binomial-distribution%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

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more