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







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$endgroup$








  • 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






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share|cite|improve this question













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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






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oldest

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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









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      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






























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