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$$ \sum_{k=1}^{15}{_NC_k}\times(0.1)^k\times (0.9)^{N-k} > 0.9996 $$

Is there a way to get $N$ from this? I was looking for a lots of materials but I couldn't find.

baeharam
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2 Answers2

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This inequality has no solutions. You want $N$ trials with success probability $0.1$ to yield either $0$ or more than $15$ successess with probability at most $0.0004$. A necessary condition for this is that the probability for $0$ successes is at most $0.0004$:

$$ 0.9^N\le0.0004\Leftrightarrow N\ge\frac{\log0.0004}{\log0.9}\approx74.3\;. $$

But at $N=74$, the probability for more than $15$ successes is already about $0.002$ (Wolfram|Alpha computation), and it grows with $N$.

joriki
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  • I cannot understand why you compute the probability for 0 successes which is at most 0.0004 – baeharam Sep 07 '18 at 10:20
  • @baeharam: Do you understand this sentence? "You want $N$ trials with success probability $0.1$ to yield either $0$ or more than $15$ successess with probability at most $0.0004$." – joriki Sep 07 '18 at 10:21
  • It means probability of more than 15 successes is less than or equal to 0.0004 – baeharam Sep 07 '18 at 10:24
  • @baeharam: What are you referring to by "it"? My sentence or your question? – joriki Sep 07 '18 at 10:27
  • My question.... – baeharam Sep 07 '18 at 10:28
  • @baeharam: It doesn't. Your sum starts at $k=1$. For it to mean what you wrote in the comment, it would have to start at $k=0$. If you want to ask the question with $k$ starting at $0$, please ask a new question rather than editing this one, as it's already got answers and changing it would cause confusion. – joriki Sep 07 '18 at 10:31
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    I posted again https://math.stackexchange.com/q/2908528/590853 – baeharam Sep 07 '18 at 11:32
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There is no closed-form expression for the $\text{cdf}$ of a Binomial law.

If you look at the sums of the first three terms for increasing $N$,

$$ p^2+2pq+q^2,\\ p^3+3p^2q+3pq^2=p(p^2+2pq+q^2)+pq(p+2q),\\ p^4+4p^3q+6p^2q^2=p(p^3+3p^2q+3pq^2)+p^2q(p+3q),\\ p^5+5p^4q+10p^3q^2=p(p^4+4p^3q+6p^2q^2)+p^3q(p+4q),\\ \cdots$$

a double recurrence appears (sum of the $M$ first terms as a function of the sum for $N-1$ and the sum of the $M-1$ first terms). So the amount of terms to be computed is of order $MN$.