Given the binomial PDF
$P(X = j) = \frac{n!}{(n-j)!x!}p^{j}(1-p)^{n-j}$
where $n$ is the number of trials, $j$ is the number of successes, $p$ is the probability of success, is there some simple solution to find $p$ given $n, j,$ and $P(X > j)$?
The specific example I'm attempting to solve is one of product reliability; say for example I have 3 satellites, of which I require a 90% chance for 2 to be functional after a certain time period; what reliability (chance of functioning after that time period) must each individual satellite have?
For 2 satellites the answer is trivial, their individual reliability must simply be $\sqrt0.9$ - is there an equivalent method for $3\choose2$ satellites?