Finding the probability that $6$ heads are obtained if a coin is tossed until there are $5$ tails [closed]












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A coin is tossed until $5$ tails. Find the probability of $6$ heads.
I tried this way.
$$P (X=11)=binom{10}{4}frac{1}{2^5} frac{1}{2^6}$$
I'm not sure. Can you help me with this?






probability







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closed as off-topic by Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy Dec 26 '18 at 19:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:44






  • 1




    $begingroup$
    The result looks fine to me. But how you define the random variable $X$?
    $endgroup$
    – callculus
    Dec 20 '18 at 3:49
















-1












$begingroup$


A coin is tossed until $5$ tails. Find the probability of $6$ heads.
I tried this way.
$$P (X=11)=binom{10}{4}frac{1}{2^5} frac{1}{2^6}$$
I'm not sure. Can you help me with this?






probability







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy Dec 26 '18 at 19:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:44






  • 1




    $begingroup$
    The result looks fine to me. But how you define the random variable $X$?
    $endgroup$
    – callculus
    Dec 20 '18 at 3:49














-1












-1








-1





$begingroup$


A coin is tossed until $5$ tails. Find the probability of $6$ heads.
I tried this way.
$$P (X=11)=binom{10}{4}frac{1}{2^5} frac{1}{2^6}$$
I'm not sure. Can you help me with this?






probability







share|cite|improve this question











$endgroup$




A coin is tossed until $5$ tails. Find the probability of $6$ heads.
I tried this way.
$$P (X=11)=binom{10}{4}frac{1}{2^5} frac{1}{2^6}$$
I'm not sure. Can you help me with this?






probability




probability






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share|cite|improve this question













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edited Dec 20 '18 at 20:03









N. F. Taussig

45k103358




45k103358










asked Dec 20 '18 at 3:39









Monika_j22Monika_j22

61




61




closed as off-topic by Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy Dec 26 '18 at 19:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy Dec 26 '18 at 19:28


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, Abcd, Shailesh, metamorphy

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:44






  • 1




    $begingroup$
    The result looks fine to me. But how you define the random variable $X$?
    $endgroup$
    – callculus
    Dec 20 '18 at 3:49


















  • $begingroup$
    Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
    $endgroup$
    – John Douma
    Dec 20 '18 at 3:44






  • 1




    $begingroup$
    The result looks fine to me. But how you define the random variable $X$?
    $endgroup$
    – callculus
    Dec 20 '18 at 3:49
















$begingroup$
Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
$endgroup$
– John Douma
Dec 20 '18 at 3:44




$begingroup$
Can you explain your attempt more clearly? You have posted an unformatted equation with undefined variables.
$endgroup$
– John Douma
Dec 20 '18 at 3:44




1




1




$begingroup$
The result looks fine to me. But how you define the random variable $X$?
$endgroup$
– callculus
Dec 20 '18 at 3:49




$begingroup$
The result looks fine to me. But how you define the random variable $X$?
$endgroup$
– callculus
Dec 20 '18 at 3:49










2 Answers
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The last toss is the fifth $T$. For a win you therefore need exactly $6$ heads and $4$ tails in the first ten tosses, and then a $T$ to finish it up. The probability for the first to happen is ${10choose4}cdot2^{-10}$. When you have reached this point the probability that you get a $T$ in the next toss is ${1over2}$. It follows that the probability $p$ of a win is given by $$p={10choose4}cdot2^{-11}=0.102539 .$$






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    0












    $begingroup$

    You need to have at least $6$ heads in the first $10$ tosses. Aside from an extra factor $frac 12$ (the exponent $5$ should be $4$) you have computed the chance of exactly $6$ heads.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The last toss is the fifth $T$. For a win you therefore need exactly $6$ heads and $4$ tails in the first ten tosses, and then a $T$ to finish it up. The probability for the first to happen is ${10choose4}cdot2^{-10}$. When you have reached this point the probability that you get a $T$ in the next toss is ${1over2}$. It follows that the probability $p$ of a win is given by $$p={10choose4}cdot2^{-11}=0.102539 .$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The last toss is the fifth $T$. For a win you therefore need exactly $6$ heads and $4$ tails in the first ten tosses, and then a $T$ to finish it up. The probability for the first to happen is ${10choose4}cdot2^{-10}$. When you have reached this point the probability that you get a $T$ in the next toss is ${1over2}$. It follows that the probability $p$ of a win is given by $$p={10choose4}cdot2^{-11}=0.102539 .$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The last toss is the fifth $T$. For a win you therefore need exactly $6$ heads and $4$ tails in the first ten tosses, and then a $T$ to finish it up. The probability for the first to happen is ${10choose4}cdot2^{-10}$. When you have reached this point the probability that you get a $T$ in the next toss is ${1over2}$. It follows that the probability $p$ of a win is given by $$p={10choose4}cdot2^{-11}=0.102539 .$$






          share|cite|improve this answer









          $endgroup$



          The last toss is the fifth $T$. For a win you therefore need exactly $6$ heads and $4$ tails in the first ten tosses, and then a $T$ to finish it up. The probability for the first to happen is ${10choose4}cdot2^{-10}$. When you have reached this point the probability that you get a $T$ in the next toss is ${1over2}$. It follows that the probability $p$ of a win is given by $$p={10choose4}cdot2^{-11}=0.102539 .$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 14:54









          Christian BlatterChristian Blatter

          176k8115328




          176k8115328























              0












              $begingroup$

              You need to have at least $6$ heads in the first $10$ tosses. Aside from an extra factor $frac 12$ (the exponent $5$ should be $4$) you have computed the chance of exactly $6$ heads.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You need to have at least $6$ heads in the first $10$ tosses. Aside from an extra factor $frac 12$ (the exponent $5$ should be $4$) you have computed the chance of exactly $6$ heads.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You need to have at least $6$ heads in the first $10$ tosses. Aside from an extra factor $frac 12$ (the exponent $5$ should be $4$) you have computed the chance of exactly $6$ heads.






                  share|cite|improve this answer









                  $endgroup$



                  You need to have at least $6$ heads in the first $10$ tosses. Aside from an extra factor $frac 12$ (the exponent $5$ should be $4$) you have computed the chance of exactly $6$ heads.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 5:12









                  Ross MillikanRoss Millikan

                  301k24200375




                  301k24200375















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