I need more info, so I'm assuming there are 10 people here. Each considers himself/herself a baseball fan or not a baseball fan. Let's say there's an equal probability of someone being a baseball fan or of not being a baseball fan, so p=½. Consider the equation (½+½)n=1. No matter what the value of n is this equation will always be true because 1n=1. So let n=10: (½+½)10=1. This is a binomial probability distribution and we can expand this (without evaluating it):
½10+10(½)9(½)+45(½)8(½)2+120(½)7(½)3+210(½)6(½)4+...
The coefficients can be calculated thus:
10, (10×9)/(1×2)=45, (10×9×8)/(1×2×3)=120, (10×9×8×7)/(1×2×3×4)=210, and so on.
The list of coefficients in order is:
1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1 (11th row of Pascal's Triangle).
Note that pnp10-n=p10 for every term.
The sum of the series is 1 because (½+½)10=1. So ½10(1+10+45+120+210+...)=1.
When we add the coefficients together we get 1024=210.
We are interested in the sum of the last three or the first three terms (because of the symmetry):
(1+10+45)(½10)=56/1024=7/128=0.0547 approx or 5.47%. This is P(8)+P(9)+P(10)=P(1)+P(2)+P(3) (in this case of equal probability, p=½).
But, suppose that, on average, in the population as a whole, 4 out of 5 people are baseball fans then 1 out of 5 people is not a baseball fan. This alters the balance and we have to consider something different, but similar, in most ways to what we just calculated. In this case though, p=⅘ and 1-p=⅕. Let q=1-p=⅕.
Now we have:
p10+10p9q+45p8q2+120p7q3+...
The first 3 terms are P(10), P(9), P(8), which is the probability that at least 8 people out of 10 randomly chosen are baseball fans.
We would have:
P(10)+P(9)+P(8)=p10+10p9q+45p8q2=⅘10+10(⅘)9(⅕)+45(⅘)8(⅕)2=0.6778=67.78%.