finding the probability for coin toss and expected number of flips required, is this the right answer for...












0












$begingroup$


A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.




  1. find the probability that the first head occurs in odd number of tosses ?

  2. find the expected number of flips required for the first head ?


the solution for number 1
$p(1 + (1-p)^2 + (1-p)^4 + ldots)
= pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$



the solution for number 2



$sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$



is this the right answer for this problem and thank you






probability statistics probability-distributions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.




    1. find the probability that the first head occurs in odd number of tosses ?

    2. find the expected number of flips required for the first head ?


    the solution for number 1
    $p(1 + (1-p)^2 + (1-p)^4 + ldots)
    = pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$



    the solution for number 2



    $sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$



    is this the right answer for this problem and thank you






    probability statistics probability-distributions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.




      1. find the probability that the first head occurs in odd number of tosses ?

      2. find the expected number of flips required for the first head ?


      the solution for number 1
      $p(1 + (1-p)^2 + (1-p)^4 + ldots)
      = pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$



      the solution for number 2



      $sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$



      is this the right answer for this problem and thank you






      probability statistics probability-distributions







      share|cite|improve this question









      $endgroup$




      A coin having probability $p$ of coming up head is to be successively flipped until the first head appears.




      1. find the probability that the first head occurs in odd number of tosses ?

      2. find the expected number of flips required for the first head ?


      the solution for number 1
      $p(1 + (1-p)^2 + (1-p)^4 + ldots)
      = pdisplaystylesum_{i=0}^infty ((1-p)^2)^i = frac{p}{1-(1-p)^2} = frac{1}{2-p}.$



      the solution for number 2



      $sumlimits_{i=1}^infty i p (1-p)^{i-1}=frac{1}{p}~~~~~~~$



      is this the right answer for this problem and thank you






      probability statistics probability-distributions




      probability statistics probability-distributions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 19:03









      green lifegreen life

      186




      186






















          2 Answers
          2






          active

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          0












          $begingroup$

          Both are correct.



          As a way to avoid the use of infinite series:



          For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$



          For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for explaining this more clear way to obtain it
            $endgroup$
            – green life
            Dec 16 '18 at 19:28



















          0












          $begingroup$

          Let $X = $ number of tosses until a head appears.



          Then $X$ ~ Geometric$(frac{1}{2})$.



          Part 1



          The probability we wish to calculate is,



          $$
          Pr(odd) =
          sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
          sum_{k=0}^{infty}{p(1 - p)^{2k}} =
          psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
          frac{p}{1 - (1-p)^{2}} =
          frac{1}{2-p}
          $$



          Part 2



          Let $Y = $ heads was seen on the first flip.



          Then $Y$ ~ Bernoulli$(p)$.



          $$
          E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
          $$



          This then reduces to the equation,
          $$
          E[X] = 1 + (1-p)E[X]
          $$
          And then solving for $E[X]$ we get
          $$
          E[X] = frac{1}{p}
          $$



          Remarks



          My calculations agree with yours, so I would conclude that your answers are correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for the explanation
            $endgroup$
            – green life
            Dec 16 '18 at 19:42













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Both are correct.



          As a way to avoid the use of infinite series:



          For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$



          For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for explaining this more clear way to obtain it
            $endgroup$
            – green life
            Dec 16 '18 at 19:28
















          0












          $begingroup$

          Both are correct.



          As a way to avoid the use of infinite series:



          For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$



          For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for explaining this more clear way to obtain it
            $endgroup$
            – green life
            Dec 16 '18 at 19:28














          0












          0








          0





          $begingroup$

          Both are correct.



          As a way to avoid the use of infinite series:



          For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$



          For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$






          share|cite|improve this answer









          $endgroup$



          Both are correct.



          As a way to avoid the use of infinite series:



          For the first note that we either win on the first roll, lose on the second, or reset. Thus, letting $Psi$ denote the answer, we have $$Psi=p+(1-p)ptimes 0+(1-p)^2Psiimplies left(1-(1-p)^2right)Psi=pimplies Psi=frac p{1-(1-p^2)}$$



          For the second note the recursive relation $$E=ptimes 1 +(1-p)times (E+1)implies pE=1implies E=frac 1p$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 19:13









          lulululu

          42.9k25080




          42.9k25080












          • $begingroup$
            thank you very much for confirming the answer and for explaining this more clear way to obtain it
            $endgroup$
            – green life
            Dec 16 '18 at 19:28


















          • $begingroup$
            thank you very much for confirming the answer and for explaining this more clear way to obtain it
            $endgroup$
            – green life
            Dec 16 '18 at 19:28
















          $begingroup$
          thank you very much for confirming the answer and for explaining this more clear way to obtain it
          $endgroup$
          – green life
          Dec 16 '18 at 19:28




          $begingroup$
          thank you very much for confirming the answer and for explaining this more clear way to obtain it
          $endgroup$
          – green life
          Dec 16 '18 at 19:28











          0












          $begingroup$

          Let $X = $ number of tosses until a head appears.



          Then $X$ ~ Geometric$(frac{1}{2})$.



          Part 1



          The probability we wish to calculate is,



          $$
          Pr(odd) =
          sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
          sum_{k=0}^{infty}{p(1 - p)^{2k}} =
          psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
          frac{p}{1 - (1-p)^{2}} =
          frac{1}{2-p}
          $$



          Part 2



          Let $Y = $ heads was seen on the first flip.



          Then $Y$ ~ Bernoulli$(p)$.



          $$
          E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
          $$



          This then reduces to the equation,
          $$
          E[X] = 1 + (1-p)E[X]
          $$
          And then solving for $E[X]$ we get
          $$
          E[X] = frac{1}{p}
          $$



          Remarks



          My calculations agree with yours, so I would conclude that your answers are correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for the explanation
            $endgroup$
            – green life
            Dec 16 '18 at 19:42


















          0












          $begingroup$

          Let $X = $ number of tosses until a head appears.



          Then $X$ ~ Geometric$(frac{1}{2})$.



          Part 1



          The probability we wish to calculate is,



          $$
          Pr(odd) =
          sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
          sum_{k=0}^{infty}{p(1 - p)^{2k}} =
          psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
          frac{p}{1 - (1-p)^{2}} =
          frac{1}{2-p}
          $$



          Part 2



          Let $Y = $ heads was seen on the first flip.



          Then $Y$ ~ Bernoulli$(p)$.



          $$
          E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
          $$



          This then reduces to the equation,
          $$
          E[X] = 1 + (1-p)E[X]
          $$
          And then solving for $E[X]$ we get
          $$
          E[X] = frac{1}{p}
          $$



          Remarks



          My calculations agree with yours, so I would conclude that your answers are correct.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for confirming the answer and for the explanation
            $endgroup$
            – green life
            Dec 16 '18 at 19:42
















          0












          0








          0





          $begingroup$

          Let $X = $ number of tosses until a head appears.



          Then $X$ ~ Geometric$(frac{1}{2})$.



          Part 1



          The probability we wish to calculate is,



          $$
          Pr(odd) =
          sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
          sum_{k=0}^{infty}{p(1 - p)^{2k}} =
          psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
          frac{p}{1 - (1-p)^{2}} =
          frac{1}{2-p}
          $$



          Part 2



          Let $Y = $ heads was seen on the first flip.



          Then $Y$ ~ Bernoulli$(p)$.



          $$
          E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
          $$



          This then reduces to the equation,
          $$
          E[X] = 1 + (1-p)E[X]
          $$
          And then solving for $E[X]$ we get
          $$
          E[X] = frac{1}{p}
          $$



          Remarks



          My calculations agree with yours, so I would conclude that your answers are correct.






          share|cite|improve this answer









          $endgroup$



          Let $X = $ number of tosses until a head appears.



          Then $X$ ~ Geometric$(frac{1}{2})$.



          Part 1



          The probability we wish to calculate is,



          $$
          Pr(odd) =
          sum_{k=0}^{infty}{Pr(X = 2k + 1)} =
          sum_{k=0}^{infty}{p(1 - p)^{2k}} =
          psum_{k=0}^{infty}{((1-p)^{2})^{k}} =
          frac{p}{1 - (1-p)^{2}} =
          frac{1}{2-p}
          $$



          Part 2



          Let $Y = $ heads was seen on the first flip.



          Then $Y$ ~ Bernoulli$(p)$.



          $$
          E[X] = E[E[X|Y]] = Pr(Y = 1)E[X|Y=1] + Pr(Y=0)E[X| Y=0] = p(1) + (1-p)(1 + E[X])
          $$



          This then reduces to the equation,
          $$
          E[X] = 1 + (1-p)E[X]
          $$
          And then solving for $E[X]$ we get
          $$
          E[X] = frac{1}{p}
          $$



          Remarks



          My calculations agree with yours, so I would conclude that your answers are correct.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 19:29









          mwbrulhardtmwbrulhardt

          11




          11












          • $begingroup$
            thank you very much for confirming the answer and for the explanation
            $endgroup$
            – green life
            Dec 16 '18 at 19:42




















          • $begingroup$
            thank you very much for confirming the answer and for the explanation
            $endgroup$
            – green life
            Dec 16 '18 at 19:42


















          $begingroup$
          thank you very much for confirming the answer and for the explanation
          $endgroup$
          – green life
          Dec 16 '18 at 19:42






          $begingroup$
          thank you very much for confirming the answer and for the explanation
          $endgroup$
          – green life
          Dec 16 '18 at 19:42




















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