How can I get $N$ trials from binomial distribution?

Question

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

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

Answer 1

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 $N$.

Answer 2

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 $MN$.