'll use the binomial approach. We need to calculate the probabilities that 9, 10 or 11
people have brown eyes. The probability that any one person has brown eyes is 0.8,
so the probability that they don't is 1 - 0.8 = 0.2. So the appropriate binomial terms are
(11 C 9)(0.8)^9*(0.2)^2 + (11 C 10)(0.8)^10*(0.2)^1 + (11 C 11)(0.8)^11*(0.2)^0 =
0.2953 + 0.2362 + 0.0859 = 0.6174, or about 61.7 %. Since this is over 50%, it
is more likely than not that 9 of 11 randomly chosen people have brown eyes, at
least in this region.
Note that (n C r) = n!/((n-r)!*r!). So (11 C 9) = 55, (11 C 10) = 11 and (11 C 0) = 1.
