Finding p in a binomial distribution when n and x is known

Question

I want to solve: $$b(2;10,p)\leq 0.1$$ Solving: $$\frac{10!}{2!8!}p^2q^8=0.1$$ $$pq^4=(\frac{0.1\cdot 2!\cdot 8!}{10!})^{0.5}, (q=1-p)$$ $$q^4-q^5=(\frac{0.1\cdot 2!\cdot 8!}{10!})^{0.5}$$ This is where the computer comes in. I just plug this into wolfram and get the answer. What I wanna know is, can you solve this fast and easy with just pen and paper? Alternatively, are there any ways of approximating the answer using,say, a normal distribution?

Answer 1

Assuming that you are concerned by $q >0$, consider that you look for the zero's of function $$f(q)=q^4-q^5-\sqrt{\frac 1 {450}}$$ The derivative $$f'(q)=4q^3-5q^4$$ cancels at $q_*=\frac 45$ for which $f(q_*)=\frac{256}{3125}-\frac{1}{15 \sqrt{2}}\approx 0.0347795$ and the second derivative test shows that this is a maximum.

Around this point, build a Taylor expansion $$f(q)=\left(\frac{256}{3125}-\frac{1}{15 \sqrt{2}}\right)-\frac{32}{25} \left(q-\frac{4}{5}\right)^2+O\left(\left(q-\frac{4}{5}\right)^3\right)$$ Ignoring the higher order terms and solving for $q$ gives $$q_\pm=\frac{480\pm\sqrt{15 \left(1536-625 \sqrt{2}\right)}}{600} $$ that is to say $0.635$ and $0.965$.

It is not extremely accurate but not so so bad and it is easily done with pen and paper.

We could do better and better building the $[2,n]$ Padé approximant of the function. This would write $$f(q)\sim\frac{f(q_*)+a^{(n)}_1(q-q_*)+a^{(n)}_2(q-q*)^2}{1+\sum_{k=1}^n b_k (q-q_*)^k}$$ and we can solve the quadratic in numerator.

Let me give you the solutions for a few values of $n$ $$\left( \begin{array}{cc} n & q_- & q_+ \\ 0 & 0.635162 & 0.964838 \\ 1 & 0.597735 & 0.934336 \\ 2 & 0.583560 & 0.940445 \\ 3 & 0.579455 & 0.939340 \end{array} \right)$$ which become quite good compared to the exact solutions (even $n=1$ is quite a nice approximation).

For $n=1$, the equation to be solved would just be $$6 f(q_*) f''(q_*)-2 f(q_*) f'''(q_*) (q-q_*)+3 f''(q_*)^2 (q-q_*)^2=0$$

Answer 2

Your equation $$q^4-q^5=\frac{\sqrt{450}}{450}$$ has three real solutions, $$q\approx -.4263734231, .5781872491, .9394903861$$