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 0.37³. Which would make the general formula B^x. Correct? For some reason this doesn't seem right.

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.