Find the probability of success in the first least 20 consecutive attempts












1












$begingroup$


The battery of a particular car brand starts with probability of $0.95$. Find the probability that the battery starts on the first at least $20$ consecutive attempts.



"success": the battery starts



"failure": the battery doesn't start



I have some issues with interpreting the at least part. If the question was given without it, then the answer would be the probability of success in the first attempt and success in the second attempt and ... and success in the 20th attempt. However, I managed to get to an answer, but I'm not completely sure about its validity: 1 - probability of failure in the first attempt and failure in the second attempt and ... and failure in the 19th attempt.



Thank you in advance!






probability







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    The battery of a particular car brand starts with probability of $0.95$. Find the probability that the battery starts on the first at least $20$ consecutive attempts.



    "success": the battery starts



    "failure": the battery doesn't start



    I have some issues with interpreting the at least part. If the question was given without it, then the answer would be the probability of success in the first attempt and success in the second attempt and ... and success in the 20th attempt. However, I managed to get to an answer, but I'm not completely sure about its validity: 1 - probability of failure in the first attempt and failure in the second attempt and ... and failure in the 19th attempt.



    Thank you in advance!






    probability







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      The battery of a particular car brand starts with probability of $0.95$. Find the probability that the battery starts on the first at least $20$ consecutive attempts.



      "success": the battery starts



      "failure": the battery doesn't start



      I have some issues with interpreting the at least part. If the question was given without it, then the answer would be the probability of success in the first attempt and success in the second attempt and ... and success in the 20th attempt. However, I managed to get to an answer, but I'm not completely sure about its validity: 1 - probability of failure in the first attempt and failure in the second attempt and ... and failure in the 19th attempt.



      Thank you in advance!






      probability







      share|cite|improve this question









      $endgroup$




      The battery of a particular car brand starts with probability of $0.95$. Find the probability that the battery starts on the first at least $20$ consecutive attempts.



      "success": the battery starts



      "failure": the battery doesn't start



      I have some issues with interpreting the at least part. If the question was given without it, then the answer would be the probability of success in the first attempt and success in the second attempt and ... and success in the 20th attempt. However, I managed to get to an answer, but I'm not completely sure about its validity: 1 - probability of failure in the first attempt and failure in the second attempt and ... and failure in the 19th attempt.



      Thank you in advance!






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      George R.George R.

      1,426411




      1,426411






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:



          $P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts. What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start. In other words,



          $P(Xge 20) $



          $$=1-P(Xle 19) = 1- q-pq-p^2q-p^3q-cdots - p^{19}q$$



          $$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-qfrac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?



            Edit:
            you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $sum_{i=0}^{infty}(0.95)^i$
            You can calculate where this series goes to.






            share|cite|improve this answer










            New contributor




            Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
              $endgroup$
              – George R.
              2 hours ago



















            1












            $begingroup$

            Let $i$ be an integer between 1 and 20.



            Probability that the car fails for the first time at the first try : $0,05.$



            Probability that the car fails for the first time at the second try : $0,95times 0,05$



            ...



            Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}times 0,05$ ($i-1$ consecutive successes followed by a failure).



            This gives a partition on the opposite event.



            Consequently, the probability you're looking for is :



            $1-0,05times(1+0,95+ldots+0,95^{19})=1-0,05frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
              $endgroup$
              – Satish Ramanathan
              1 hour ago










            • $begingroup$
              You're right. I edited my answer consequently.
              $endgroup$
              – Ayoub
              1 hour ago











            Your Answer





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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

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            2












            $begingroup$

            Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:



            $P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts. What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start. In other words,



            $P(Xge 20) $



            $$=1-P(Xle 19) = 1- q-pq-p^2q-p^3q-cdots - p^{19}q$$



            $$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-qfrac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:



              $P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts. What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start. In other words,



              $P(Xge 20) $



              $$=1-P(Xle 19) = 1- q-pq-p^2q-p^3q-cdots - p^{19}q$$



              $$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-qfrac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:



                $P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts. What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start. In other words,



                $P(Xge 20) $



                $$=1-P(Xle 19) = 1- q-pq-p^2q-p^3q-cdots - p^{19}q$$



                $$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-qfrac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$






                share|cite|improve this answer











                $endgroup$



                Let us define success as start and failure as non-start. The probability that you have n consecutive successes before a failure follows geometric distribution as below:



                $P(X=k) = p^kq$, $k = 0,1,2,..$where $p$ is the probability of start and $ q$ is the probability of non start and k is the number of starts. What you have been asked is to find the probability of atleast $20$ consecutive starts before a non-start. In other words,



                $P(Xge 20) $



                $$=1-P(Xle 19) = 1- q-pq-p^2q-p^3q-cdots - p^{19}q$$



                $$ 1- q(1+p+p^2+..p^{19})$$ $$= 1-qfrac{(1-p^{20})}{1-p} = 1-1+p^{20} = p^{20}=0.358486$$ where $p = 0.95$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                Satish RamanathanSatish Ramanathan

                9,59031323




                9,59031323























                    1












                    $begingroup$

                    If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?



                    Edit:
                    you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $sum_{i=0}^{infty}(0.95)^i$
                    You can calculate where this series goes to.






                    share|cite|improve this answer










                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                      $endgroup$
                      – George R.
                      2 hours ago
















                    1












                    $begingroup$

                    If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?



                    Edit:
                    you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $sum_{i=0}^{infty}(0.95)^i$
                    You can calculate where this series goes to.






                    share|cite|improve this answer










                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                      $endgroup$
                      – George R.
                      2 hours ago














                    1












                    1








                    1





                    $begingroup$

                    If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?



                    Edit:
                    you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $sum_{i=0}^{infty}(0.95)^i$
                    You can calculate where this series goes to.






                    share|cite|improve this answer










                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    If you are looking for the probability that at least one of the first 20's attempts will make the battery start, you are looking for 1 - "all the 20's will fail". is it making it any easier for you? Am I getting your question right?



                    Edit:
                    you can take the probability for all first 20 successes (0.95^20) and multiply it in all your options, which is the geometric series $sum_{i=0}^{infty}(0.95)^i$
                    You can calculate where this series goes to.







                    share|cite|improve this answer










                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago





















                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 2 hours ago









                    Shahar MaziaShahar Mazia

                    134




                    134




                    New contributor




                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Shahar Mazia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.












                    • $begingroup$
                      I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                      $endgroup$
                      – George R.
                      2 hours ago


















                    • $begingroup$
                      I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                      $endgroup$
                      – George R.
                      2 hours ago
















                    $begingroup$
                    I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                    $endgroup$
                    – George R.
                    2 hours ago




                    $begingroup$
                    I am looking for the probability that in at least 20 of the first tries only successes occur, that is: all first 20 attempts are successful or all first 21 attempts are successful or ... .
                    $endgroup$
                    – George R.
                    2 hours ago











                    1












                    $begingroup$

                    Let $i$ be an integer between 1 and 20.



                    Probability that the car fails for the first time at the first try : $0,05.$



                    Probability that the car fails for the first time at the second try : $0,95times 0,05$



                    ...



                    Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}times 0,05$ ($i-1$ consecutive successes followed by a failure).



                    This gives a partition on the opposite event.



                    Consequently, the probability you're looking for is :



                    $1-0,05times(1+0,95+ldots+0,95^{19})=1-0,05frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                      $endgroup$
                      – Satish Ramanathan
                      1 hour ago










                    • $begingroup$
                      You're right. I edited my answer consequently.
                      $endgroup$
                      – Ayoub
                      1 hour ago
















                    1












                    $begingroup$

                    Let $i$ be an integer between 1 and 20.



                    Probability that the car fails for the first time at the first try : $0,05.$



                    Probability that the car fails for the first time at the second try : $0,95times 0,05$



                    ...



                    Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}times 0,05$ ($i-1$ consecutive successes followed by a failure).



                    This gives a partition on the opposite event.



                    Consequently, the probability you're looking for is :



                    $1-0,05times(1+0,95+ldots+0,95^{19})=1-0,05frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                      $endgroup$
                      – Satish Ramanathan
                      1 hour ago










                    • $begingroup$
                      You're right. I edited my answer consequently.
                      $endgroup$
                      – Ayoub
                      1 hour ago














                    1












                    1








                    1





                    $begingroup$

                    Let $i$ be an integer between 1 and 20.



                    Probability that the car fails for the first time at the first try : $0,05.$



                    Probability that the car fails for the first time at the second try : $0,95times 0,05$



                    ...



                    Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}times 0,05$ ($i-1$ consecutive successes followed by a failure).



                    This gives a partition on the opposite event.



                    Consequently, the probability you're looking for is :



                    $1-0,05times(1+0,95+ldots+0,95^{19})=1-0,05frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$






                    share|cite|improve this answer











                    $endgroup$



                    Let $i$ be an integer between 1 and 20.



                    Probability that the car fails for the first time at the first try : $0,05.$



                    Probability that the car fails for the first time at the second try : $0,95times 0,05$



                    ...



                    Probability that the car fails for the first time at the $i$-th try : $0,95^{i-1}times 0,05$ ($i-1$ consecutive successes followed by a failure).



                    This gives a partition on the opposite event.



                    Consequently, the probability you're looking for is :



                    $1-0,05times(1+0,95+ldots+0,95^{19})=1-0,05frac{1-0,95^{20}}{1-0,95}=1-(1-0,95^{20})=0,95^{20}$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 1 hour ago









                    AyoubAyoub

                    765




                    765












                    • $begingroup$
                      20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                      $endgroup$
                      – Satish Ramanathan
                      1 hour ago










                    • $begingroup$
                      You're right. I edited my answer consequently.
                      $endgroup$
                      – Ayoub
                      1 hour ago


















                    • $begingroup$
                      20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                      $endgroup$
                      – Satish Ramanathan
                      1 hour ago










                    • $begingroup$
                      You're right. I edited my answer consequently.
                      $endgroup$
                      – Ayoub
                      1 hour ago
















                    $begingroup$
                    20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                    $endgroup$
                    – Satish Ramanathan
                    1 hour ago




                    $begingroup$
                    20 consecutive starts before the first failure happens. I think you have understood the problem wrongly. In system reliability, they want you to find the probability that it starts without a trouble for alteast 20 consecutive times.
                    $endgroup$
                    – Satish Ramanathan
                    1 hour ago












                    $begingroup$
                    You're right. I edited my answer consequently.
                    $endgroup$
                    – Ayoub
                    1 hour ago




                    $begingroup$
                    You're right. I edited my answer consequently.
                    $endgroup$
                    – Ayoub
                    1 hour ago


















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                    Metroo de Marsejlo

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                    Metroo de Marsejlo Logoo Ĝeneralaj informoj Loko Marsejlo  Francio Transportsistema speco Metroo Komenco de trafiko 26-a de novembro 1977 Teknikaj informoj Longeco de sistemo 21,6 km Nombro de linioj 2 Nombro de veturiloj 36 Nombro de stacioj 28 (+ 2 interkonektoj) Aliaj informoj Traklarĝo 1 435 mm Transportisto RTM v   •   d   •   r La Metroo de Marsejlo (Francio) (france  Métro de Marseille ) estas metrosistemo funkcianta per pneŭoj derivita de la teknologio disvolvigita de la RATP por la metroo de Parizo. Ĝi malfermiĝis unuafoje al la publiko en la jaro 1977. Ĝi konsistas el du linioj, parte subteraj, kun longeco de entute 21,6 km , situantaj nur en la municipo Marsejlo. Nuntempe konstruiĝas du plilongigaĵoj. Enhavo 1 Mapo 2 Metrolinioj 2.1 Linio 1 2.2 Linio 2 3 Vidu ankaŭ 4 Eksteraj ligiloj Mapo | Mapo de la metroo de Marsejlo Metrolinioj | ...
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