Poisson distribution - find value for λ given a known probability I am trying to solve the following:
The number of particles decaying per second in a radioactive material with a high half-life (e.g. of several millennia) is (in very good approximation) Poisson-distributed.
a) On average, 20.00 particles per second decay from a radioactive sample. What is the probability that 20 particles decay in one second?
b) An average of 20.00 particles per second decay from a radioactive sample. What is the probability that at least 10 particles decay in one second?
c) From a radioactive sample, it is known that with a probability of 1%, at most 3 particles per second decay. How many particles in this period decay on average in one second? Give your result to three decimal places.

My Solution:
a) $$p(x=20)=\frac{20^{20}}{20!}e^{-20} = 0.089$$
b) $$p(x \geq 10)= 1- p(x \leq 9) \\
           = 1- (\sum\limits_{x=0}^{9} \frac{20^{x}}{x!}e^{-20}) \\
           = 1- 0.005   \\
           = 0.995
$$
c) $$p(x \leq 3)=  \sum\limits_{x=0}^{3} \frac{\lambda ^{x}}{x!}e^{-\lambda} =0.01 \\
          0.01 = (\frac{\lambda^0}{0!} +\frac{\lambda^1}{1!}+ \frac{\lambda^2}{2!} +\frac{\lambda^3}{3!}) e^{-\lambda}   \\
           = (1 +\lambda+ \frac{\lambda^2}{2} +\frac{\lambda^3}{6}) e^{-\lambda}   \\
= \frac{(\lambda^3 +3\lambda^2 + 6\lambda+ 6)} {6e^{\lambda}}   \\
$$
Using an online calculator, I found out that $\lambda \approx 10.0451$, but I'm not entirely sure whether that's the right answer.
My question is:
If my calculations are correct so far: How to calculate $\lambda$?
 A: As said, your formulation is very correct and you end with the problem of finding the zero of function $$f(\lambda)=\frac{\lambda^3 +3\lambda^2 + 6\lambda+ 6} {6e^{\lambda}}-0.01$$ If you made a plot of it, you more than likely noticed that you need to zoom more and more to locate an approximate solution.
Suppose now that you change the problem to : find the zero of function
$$g(\lambda)=\log\left(\frac{\lambda^3 +3\lambda^2 + 6\lambda+ 6} {6e^{\lambda}}\right)-\log(0.01)$$ Plot it and notice how easier it is to be and how clear it is that the solution is around $10$.
Now, let us use Newton method as @heropup recommended. Starting with the same estimate, the iterates will  be
$$\left(
\begin{array}{cc}
 n & \lambda_n \\
 0 & 15.0000 \\
 1 & 10.2623 \\
 2 & 10.0458 \\
 3 & 10.0451
\end{array}
\right)$$
A: Yes, your calculations are correct.  The last part of the question is not possible to solve in closed form; only numeric methods will work.  One way is via recursive bisection; e.g., guess-and-check.  Another way is to use Newton's method to obtain a convergent recurrence for the root; e.g., we want to find the unique positive real root of $f(z) = z^3 + 3z^2 + 6z + 6 - \frac{6}{100}e^z$, from which we obtain the recurrence $$z_{n+1} = z_n - \frac{f(z_n)}{f'(z_n)} = z_n - 1 - \frac{50z^3}{3(e^{z_n} - 100 - 100z_n - 50z_n^2)}.$$  This requires an intelligent choice of initial guess.  Clearly, $\lambda$ must be larger than $3$, since $\Pr[X \le 3] = 0.01$.  But when in part (b) we used $\lambda = 20$ we found $\Pr[X \le 9] \approx 0.005$, which is too small.  So it is natural to try something like $\lambda = 15$.  Setting $z_0 = 15$, we obtain the iterates
$$\begin{array}{c|c}
n & z_n \\ 
\hline 
0 & 15. \\
1 & 14.0173 \\
2 & 13.0551 \\
3 & 12.1362 \\
4 & 11.3037 \\
5 & 10.6312 \\
6 & 10.2101 \\
7 & 10.0612 \\
8 & 10.0453 \\
9 & 10.0451 \\
10 & 10.0451 \\
\end{array}$$
after which point we have roughly the correct answer.  Substituting into the original equation confirms that it works; e.g., $$\frac{(10.0451)^2 + 3(10.0451)^2 + 6(10.0451) + 6}{6e^{10.0451}} \approx 0.01.$$
