Those not having high serum cholesterol is 92%, so the chance of all 10 not having the high level is 0.92^10=0.4344 or 43.44%.
Let's start with the first two out of the 10 having a high level: 0.08^2 followed by 8 not having a high level=0.08^2*0.92^8=0.0032846. However, there are 45 ways (=10*9/2!) of selecting 2 women out of 10, so the probability of any combination of 2 out of 10 having a high level is 45*0.0032846=0.1478=14.78%.
The probability of at least 2 means the sum of exactly 2, 3, 4,...10 having a high level. for p=0.08 the binomial distribution is (p+(1-p))^10=1, which is the sum of all possibilities, which is obviously 100%. The expansion of the this is: p^10+10p^9*(1-p)+45p^8*(1-p)^2+...+45p^2(1-p)^8+10p(1-p)^9+(1-p)^10. The first term is the probability of all 10 women having a high level, the last term is the probability of no women having a high level. The second term is the probability of 9 out of 10 with a high level; the penultimate term is the probability of just one with a high level. The third term is the probability of 8 out of 10 with a high level; the pre-penultimate term is the probability of just 2 with a high level. The sum of the first 3 terms is the probability of at most 2 with a high level; the sum of the last 3 terms is the probability of at most 2 without a high level. Therefore, the probability of at least 2 having a high level is (1-probability of none with a high level plus probability of just 1 with a high level). This probability=1-10*0.08*0.92^9+0.92^10=0.1879=18.79%.
The probability of exactly 1 with a high level is 10p(1-p)^9=0.3777=37.77%.