Before beginning, let's lay out a few ground assumptions. For now, we're working with 50-50 chance of each of the events happening. We're also assuming that all of the events are independent. That is to say, the outcome of events 2 and 3 are in no way affected by the outcome of event 1.
With that out of the way, let's now look at the simplest possible case, where there's just one event. It can either happen or not, so there's two outcomes. That means the probability of 0 "successes" is 1/2, as is the probability of 1 success. Now what about two events? If the first event happens, then the second can either happen or not. If the first event doesn't happen, then the second can, again, either happen or not. That makes 4 total outcomes. In order to get 0 successes, both events must not happen. That occurs with probability 1/2 * 1/2 = (1/2)^2 = 1/4. For one success, either the first
or the second must happen, and the other not happen. Individually, we know that the probability of that would also be 1/2 * 1/2 = 1/4. But there's two possible arrangements so the total probability is 1/4 + 1/4 = 1/2. For two successes, both events must happen, so that gives probability 1/2 * 1/2 = 1/4.
So, then we come to the case of three events. Similar logic can show that to get 0 successes, all three events must not happen. That gives probability 1/2 * (1/2 * 1/2) = 1/2 * (1/2)^2 = (1/2)^3 = 1/8. To get 1 success, any one of the three events must happen, and the other two not. This occurs with probability 1/8. But we have 3 possible events and we must choose 1 of them, so there's three ways it can happen. Thus the total probability is 3/8. To get 2 successes, two must happen and two not. Individually, this also has probability 1/8. But we have 3 events can we need to choose 2 of them. For now, we can just count the ways easily, but this will get crazy if we had 17 events and wanted to pick 9, for instance. So we need a better way. As it turns out, there's a handy formula we can use to save time,
the binomial coefficienthttp://www.mathwords.com/b/binomial_coefficients.htm. Using that reveals that "3 choose 2" is also 3, so two successes also has probability 3/8. To get 3 successes, all the events must not happen. That has probability 1/8. This all matches the results you got.
At this point, the question becomes: Are you seeing a pattern? If not, perhaps try seeing what happens when there's four events. Once that's taken care of, all that's left to figure out is what happens if the probability
isn't 50-50? We can use a very general case and say the probability of success is
p. That, of course, means the probability of failure is
1 - p. Returning to the case with three events...
n successes: (probability of outcome) * (number of outcomes)
-----
0 successes: (p)^0 * (1-p)^3 * (3 choose 0)
1 success: (p)^1 * (1-p)^2 * (3 choose 1)
2 successes: (p)^2 * (1-p)^1 * (3 choose 2)
3 successes: (p)^3 * (1-p)^0 * (3 choose 3)
It should be noted that, due to the definition of the binomial coefficient, and the definition of the factorial, 0! = 1, and so (n choose 0) is always 1.
