# Probability >=1 Event, Multiple Independent Binomial Trials with Differing Probabilities

I apologise in advance for what is likely a very simple question, however I have tried to find the answer on here (as well as via alt sources) and I am still struggling for a clear explanation that I can extrapolate/understand.

Here is the setup and assumptions:

Three independent trials are run (A,B,C). The expected probability of an event occurring is 0.7, 0.5 and 0.2 for trials A,B and C respectively. What is the probability that over the three trials >=1 event will occur, how is this determined and how can it be extrapolated to determine the same for N trials, each with varying event probabilities.

My understanding so far:

Trial A: P = 0.7, 1-P = 0.3

Trial B: P = 0.5, 1-P = 0.5

Trial C: P = 0.2, 1-P = 0.8

All Trials yield an Event : P = 0.7 * 0.5 * 0.2 = 0.07

No Trial yields an Event : P = 0.3 * 0.5 * 0.8 = 0.12

I know that with N independent trials of the same event probability the answer is 1-((1-P)^N) - like in the coin toss example frequently cited on here

However, I'm pretty sure in this case the answer is not 1 - 0.12 = 0.88 (i.e. 1 - cumulative P of no events across the trials)

The below post (see link) appears to answer this, however I'm still not clear on the approach and answer (and evidently not a stat guy....). Thank you in advance!

Probability distribution for different probabilities

## 1 Answer

Actually, the answer is $1 - 0.12 = 0.88$; this follows as a consequence of the negation law of probability. In general, if you have $N$ independent trials with binary outcomes $X_i \sim \text{Bern}(1-p_i)$ (so $p_i$ is the probability of the event not occurring), and you let $M = \sum_{i=1}^N X_i$ be the total number of events that occur in these $N$ trials, then you have:

$$\mathbb{P}(M \geqslant 1) = 1 - \mathbb{P}(M=0) = 1 - \prod_{i=1}^N \mathbb{P}(X_i=0) = 1 - \prod_{i=1}^N (1-p_i).$$

In the case of equal probabilities $p = p_1 = ... = p_N$ you get $\mathbb{P}(M \geqslant 1) = 1 - (1-p)^N$.