# How to calculate $p_c$ (binomial probability)

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

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

$$p_c=0.00321898475092782$$

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.