My name's Chris. Level of question: ? I'm a college grad. I have been searching for the answer for this for days and have been unable to find a satisfactory solution: There are two possible outcomes for a random event, A and B. The probability of A being the outcome is 63%, and B 37%. What is the likelihood that B will be chosen twice, consecutively? Three times, ten? Here's what I'm thinking: There is a 0.37 chance that B will be chosen once. So, if B is chosen, there will be a .37 chance that it will be chosen again, or 37% of 37%. So, the likelihood that B will be chosen is 0.37 * 0.37. or .1369. Now, if B is chosen three times, then there is a 0.37 chance the first time, 0.37 the second time, 0.37 the third, or 0.37 * 0.37 * 0.37 or 0.37^{3}. Which would make the general formula B^{x}. Correct? For some reason this doesn't seem right. Chris Hi Chris,You are completely right. The probability that event B will happen X times consequently is (0.37)^{X}. If you want percents, multiply the answer by 100. Andrei
