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