Are you studying binomials? You have a medicine that has an (4/5) chance of working with any particular individual, and 16 possible outcomes: 15 improve, 14 improve, 13 improve ... 1 improves, 0 improves. There is only one combination that will result in 15 improving: (4/5)^15. Likewise, there is only one combination that will result in 0 improving: (1/5)^15. In between, it gets hairy. For example, there are 15 ways for only 1 person to improve, and you have to account for each one. In other words: A improves, B-O do not; B improves, A & C-O do not, etc. Thus, the odds that only 1 person will improve is 15 * (4/5) * [(1/5^14)]. When you start looking at the number of ways for only 2 persons to improve, only 3 persons to improve, etc., there are lots and lots of permutations; and you have to count them all. Generally, you are looking for the coefficients for each possible combination, i.e. the number of permutations that make that combination. The good news is, there's a shortcut. You can determine the coefficient of each combination using the formula {n! / [k! (n-k)!]}. This formula tells you how many ways you can have k unordered outcomes from n possibilities, i.e. the number of permutations that form the combination where k people out of n improve. Thus, the number of ways you can have 15 people improve is 15! / (0! 15!) = 1, the number of ways you can have 0 people improve is 15! / (15! 0!) = 1, and the number of ways you can have just 1 person improve is 15! / (1! 14!) = 15, just as above. There are 16 possible combinations, and the number of ways each can be made is discoverable using this formula. PROBLEM 1: What is the probability that fewer than 9 show improvement? This will be the sum of the probabilities of nine different combinations occurring: 0 people improving, 1 person improving, 2 people improving ... 8 people improving. That will be ((15!/(0!*15!)) * ((4/5)^0)*((1/5)^15)) + ((15!/(1!*14!)) * ((4/5)^1)*((1/5)^14)) + ... + ((15!/(8!*7!)) * ((4/5)^8)*((1/5)^7)). I constructed an Excel spreadsheet to calculate each case for me; the sum of these nine probabilities is approximately 1.8%. PROBLEM 2: What is the probability that 12 or more show improvement? This will be the sum of the probabilities of four different combinations occuring: 12 people improving, 13 people improving, etc. That will be ((15!/(12!*3!)) * ((4/5)^12)*((1/5)^3)) + ... + ((15!/(15!*0!)) * ((4/5)^15)*((1/5)^0)). The sum of these four probabilities is approximately 64.8%. The general formula for the probability of a given number of people improving is (15!/(x!(15-x)!))*((4/5)^x)*((1/5)^(15-x)).