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I would like to isolate p in the following. I am not sure if it is even possible.

a = B(1; 10, p)

B(x;n,p) is the cumulative binomial function.

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    If I understand you correctly, then you have (expanding the formula) $a = (1-p)^{10}+10 p (1-p)^9$. For$a$given $a$ and for this particular expression, there may be 0, 1 or 2 solutions (ie, values of$p$that satisfy the expression).2012-05-18

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The aim is to solve $a=u(p)$ with $u(p)=(1-p)^{10}+10p(1-p)^9=(1+9p)(1-p)^9$. Since $u$ is a high degree polynomial, there can exist no formula inverting it in full generality and using only usual functions. However...

Since $u'(p)\lt0$ for every $p$ in $(0,1)$, the function $u$ is decreasing on $[0,1]$ from $u(0)=1$ to $u(1)=0$. Hence, for each $a$ in $(0,1)$, there exists a unique value $p_a$ in $(0,1)$ such that $u(p_a)=a$.

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    @Norham: you could construct a numerical method for solving $p$, seeing that this is for a program... Newton-Raphson + bisection should do fine.2012-05-18