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 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 $P(X\geq 3)=10.175\%$
