enter image description here In the given image I don't understand how to calculate $p_c$, can anyone please explain?

enter image description here

  • $\begingroup$ Do you mean the whole derivation or just how to calculate $p_c$ from the equation in the last line? $\endgroup$ – Matt L. Nov 19 '18 at 17:07
  • $\begingroup$ just the Pc in the last line $\endgroup$ – Cynthia Nov 19 '18 at 17:17
  • $\begingroup$ I am stuck because I got two Pc with different power here $\endgroup$ – Cynthia Nov 19 '18 at 17:23

Just as in your previous question, there is no analytic solution to that problem. You're required to find some quick approximation that gives you a useful result. Rewrite the last equation as

$$p_c=\sqrt[9]{\frac{2.79\cdot 10^{-8}}{\binom{204}{9}}}\frac{1}{(1-p_c)^{195/9}}\tag{1}$$

If we boldly assume that $p_c\ll 1$ holds, we can use the following approximation:

$$\frac{1}{(1-p_c)^{195/9}}\approx 1\tag{2}$$

which, when combined with $(1)$, results in

$$p_c\approx \sqrt[9]{\frac{2.79\cdot 10^{-8}}{\binom{204}{9}}}\approx 0.003002\tag{3}$$

The exact numerical solution is


which shows that the approximation $(3)$ is reasonably good.


You can use the binomial approximation: $$ (1 + x)^\alpha \approx 1 + \alpha x, $$ which should be a very good approximation. The conditions for this being accurate are that $|\alpha x| \ll 1$.

When you substitute this expression into your total probability, you now have a polynomial you can solve for the roots of.


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