I'd like to calculate the probability that at least N independent events out of a set of M events complete successfully. Each event Ei has a probability of success of Pi.

For example, if there are four independent events each with a unique probability of success:

E1: P1

E2: P2

E3: P3

E4: P4

How do I calculate that probability that at least 1 event will be successful? At least 2 events? 3 events? The order of successes does not matter. Thanks for any help!

The best answer I could find is here:


Unfortunately for me, it doesn't look like there's an easy way to get rid of the looping (summation) to solve the problem.

  • 2
    $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Oct 19, 2016 at 1:18
  • $\begingroup$ This is not a course problem, it's actually a business problem (flight scheduling). I'm trying to calculate the probability of at least M of N flights being completed given that each flight has an individual probability of completion. I see how I can get exactly M of N events using binomial series, and I suppose that I can just add up M, M+1, ... N probabilities in a loop to get the overall probability. I was hoping there was a way to simplify the expression to one formula and not use a loop for computational efficiency. $\endgroup$
    – dmjones
    Oct 21, 2016 at 3:14

1 Answer 1


The probability that no event occurs is$$\prod_i (1-p_i)$$hence the probability that at least one event occurs is$$1-\prod_i (1-p_i)$$The probability that exactly one event occurs is$$\sum_{j=1}^m p_j\prod_{i\ne j}(1-p_i)$$and so on. The probability of exactly $k$ events is thus $$\sum_{1\le i_1<i_2<\cdots<i_k\le m} \prod_{j=1}^k p_{i_j} \prod_{j=k+1}^m (1-p_{i_j})$$where the second product is on the indices that are not in the first group $\{i_1,\ldots,i_k\}$


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