How can I get N trials from binomial distribution (Edited)











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$$
sum_{k=0}^{17}{_NC_k}times 0.1^ktimes 0.9^{N-k}<0.004
$$



How can I get a $N$ from above inequality?










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  • If $N=17$ the LHS is $1$...
    – gt6989b
    Sep 7 at 11:39










  • $N$ is not 17, I don't know $N$
    – baeharam
    Sep 7 at 11:42










  • This answer to this related (but distinct) question is relevant.
    – joriki
    Sep 7 at 12:16










  • I cannot understand what he says, what is it about?
    – baeharam
    Sep 7 at 12:22















up vote
0
down vote

favorite












$$
sum_{k=0}^{17}{_NC_k}times 0.1^ktimes 0.9^{N-k}<0.004
$$



How can I get a $N$ from above inequality?










share|cite|improve this question






















  • If $N=17$ the LHS is $1$...
    – gt6989b
    Sep 7 at 11:39










  • $N$ is not 17, I don't know $N$
    – baeharam
    Sep 7 at 11:42










  • This answer to this related (but distinct) question is relevant.
    – joriki
    Sep 7 at 12:16










  • I cannot understand what he says, what is it about?
    – baeharam
    Sep 7 at 12:22













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$
sum_{k=0}^{17}{_NC_k}times 0.1^ktimes 0.9^{N-k}<0.004
$$



How can I get a $N$ from above inequality?










share|cite|improve this question













$$
sum_{k=0}^{17}{_NC_k}times 0.1^ktimes 0.9^{N-k}<0.004
$$



How can I get a $N$ from above inequality?







binomial-distribution






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share|cite|improve this question











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share|cite|improve this question










asked Sep 7 at 11:32









baeharam

587




587












  • If $N=17$ the LHS is $1$...
    – gt6989b
    Sep 7 at 11:39










  • $N$ is not 17, I don't know $N$
    – baeharam
    Sep 7 at 11:42










  • This answer to this related (but distinct) question is relevant.
    – joriki
    Sep 7 at 12:16










  • I cannot understand what he says, what is it about?
    – baeharam
    Sep 7 at 12:22


















  • If $N=17$ the LHS is $1$...
    – gt6989b
    Sep 7 at 11:39










  • $N$ is not 17, I don't know $N$
    – baeharam
    Sep 7 at 11:42










  • This answer to this related (but distinct) question is relevant.
    – joriki
    Sep 7 at 12:16










  • I cannot understand what he says, what is it about?
    – baeharam
    Sep 7 at 12:22
















If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39




If $N=17$ the LHS is $1$...
– gt6989b
Sep 7 at 11:39












$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42




$N$ is not 17, I don't know $N$
– baeharam
Sep 7 at 11:42












This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16




This answer to this related (but distinct) question is relevant.
– joriki
Sep 7 at 12:16












I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22




I cannot understand what he says, what is it about?
– baeharam
Sep 7 at 12:22










1 Answer
1






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1
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As commented by @joriki, in general for this summation a numerical approach is the way to go.



Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.



The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.



Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2{500}$ with $p=0.1 = frac1{10}$) is beyond me, and I'd like to hear from anyone who has an idea.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    As commented by @joriki, in general for this summation a numerical approach is the way to go.



    Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.



    The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.



    Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2{500}$ with $p=0.1 = frac1{10}$) is beyond me, and I'd like to hear from anyone who has an idea.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      As commented by @joriki, in general for this summation a numerical approach is the way to go.



      Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.



      The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.



      Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2{500}$ with $p=0.1 = frac1{10}$) is beyond me, and I'd like to hear from anyone who has an idea.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        As commented by @joriki, in general for this summation a numerical approach is the way to go.



        Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.



        The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.



        Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2{500}$ with $p=0.1 = frac1{10}$) is beyond me, and I'd like to hear from anyone who has an idea.






        share|cite|improve this answer














        As commented by @joriki, in general for this summation a numerical approach is the way to go.



        Courtesy of Wolfram Alpha, after some quick trial and error one finds that at $N = 305$ the summation is about 0.00405, and at $N = 306$ the sum is roughly 0.00385.



        The minimal $N$ that satisfies your inequality is $N = 306$, which happens to be a multiple of 17.



        Whether there's an analytic solution for this particular set of numbers ($17$ and $0.004 = frac2{500}$ with $p=0.1 = frac1{10}$) is beyond me, and I'd like to hear from anyone who has an idea.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 5:26

























        answered Sep 7 at 13:13









        Lee David Chung Lin

        3,47031038




        3,47031038






























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