Binomial distribution practice problem (probability medicine will help patient, p = 0.9) The probability that a person will be helped by a certain medicine is 0.90. A doctor will be seeing 12 patients today. Determine the following: a) P(The medicine will help exactly 10 patients) b) P(The medicine helps all of the first 10 people) For (a), I got (0.9)^10 * (0.1)^2 * 12!/10!2! for (b), I got (0.9)^10 * (0.1)^2 Is that right? Or are the last two patients irrelevant for part (b)? I.e., would the probability actually be (0.9)^10 for part b? Thanks kindly.

The last two patients are indeed irrelevant for (b). The answer is $0.9^{10}$.