# Probability of at least one success in a series of independent, non-identical Bernoulli trials

Let's say I have a set of independent Bernoulli trials each with a different probability:

$$x_i \sim \operatorname{Bernoulli}(p_i)$$

The number of successes (sum of x) will be distributed according to the Poisson-Binomail distribution:

$$S = \sum_{i = 0}^n x_i \sim \operatorname{PoissonBinomial}(p)$$

However, I am interested in the the probability that there is at least 1 success. In other words:

$$S > 0 \sim \cdots\text{?}$$

• $S>0$ has a distribution. $p(S>0)$ can be calculated, but does not itself have a distribution. Which are you asking? – jsk Jun 20 at 17:26
• Sorry, yes what I'm interested in is the distribution of $S > 0$. Thank you – snakeoilsales Jun 20 at 17:34
• S>0 is a binary event (yes, no) so the distribution for it is Bernoulli. – Tim Jun 20 at 18:09
• The distribution of $1 - \sum_i (1 - p_i)$ depends on the distribution of the $p$s. If the $p$s are known exactly, then the distribution of $1 - \sum_i (1 - p_i)$ is a delta function. – Bridgeburners Jun 20 at 18:47
• S=0 means all the $x_i$ are 0, so why are you summing $1-p_i$? – jsk Jun 20 at 19:55

The probability that there is at least one success is P(S>0).

Because there can't be less than S successes and S is a discrete quantity, then: P(S>0) = 1 - P(S=0)

Each trial has a probability of success p_i, so it has a probability of failure (1-p_i)

P(S=0) = P(all fail)

Because the n trials are independent, then P(all fail) = P(trial 1 fails) * ... * P(trial n fails)

And those are the (1-p_i), so:

P(S=0) = (1-p_1) * ... * (1-p_n)