The total probability of a minimum amount probability-varied choosers choosing one option over another. OK, the title may seem a bit convoluted but here it goes. This is an simplified example of what I am trying to get at. I am really more interested in the method of figuring this out, and how it can be applied to other, similar, problems.
10 people are presented with a binary choice with Option A and Option B. Person 1 has a 25% chance of picking Option B, Person 2 has a 30% chance, and so on, incriminating 5% each person (for simplicity sake). What is the total probability, in percentage, that at least 3 of these people will pick Option B?
 A: An exact answer is going to be painful. Let $p_i$ be the probability that person $i$ chooses box B, and these probabilities are independent.
Let $Q=(1-p_1)\cdots(1-p_{10})$. 
Then the probability that at least three of them pick $B$ is:
$$1-Q\left[1+\sum_{i=1}^{10} \frac{p_i}{1-p_i} + \sum_{1\leq i<j\leq 10}\frac{p_i}{1-p_i}\frac{p_j}{1-p_j}\right]$$ 
In your case, $p_i=\frac{1}{4}+\frac{i-1}{20}=\frac{4+i}{20}$, so 
$$\begin{align}
Q&=\frac{14!}{4!20^{10}}\\\frac{p_i}{1-p_i}&=\frac{4+i}{16-i}
\end{align}$$
That sum is almost certainly still going to be complicated to calculate.
This value is just subtacting the cases where exactly zero, one, or two of the people select $B$.
The number of people choosing exactly three is:
$$Q\left[\sum_{1\leq i<j<k\leq 10} \frac{p_i}{1-p_i}\frac{p_j}{1-p_j}\frac{p_k}{1-p_k}\right]$$
Again, ugly.
A: A smaller example to show how the calculation works. Suppose that you have $4$ people. Then 
$P(X\geq 3)=P(X=3)+P(X=4)$
$P(X=3)=0.25\cdot 0.3\cdot 0.35\cdot (1-0.4)$
$+0.25\cdot 0.3\cdot (1-0.35)\cdot 0.4$
$+0.25\cdot (1-0.3)\cdot 0.35\cdot 0.4$
$+(1-0.25)\cdot 0.3\cdot 0.35\cdot 0.4$
$P(X=4)=0.25\cdot 0.3\cdot 0.35\cdot 0.4$
Thus $P(X\geq 3)=10.175\%$
