Number of tries until Failure with n different independent Bernoulli experiment I have 3(n) coffee machines in an office. I have a historical log of these machines, and I know in the last ten days(t) how many times they failed to make a coffee.
For example:

*

*machine 1: 10 failures in 100 tries

*machine 2: 20 failures in 1000 tries

*machine 3: 50 failure in 300 tries

I know that I can use the Expected Value of the Geometric distribution for a machine to calculate the mean number of tries before the first failure.
Question:
How many times does one person need to use coffee machines in this office until he gets the first failure?
How can we formulate this problem?

As noticed in the comment by @periwinkle, the result depends also on the probability of the machine by a user. I would assume that the machines are chosen uniformly at random, but any other probability distribution can be used instead. Recall the law of total probability
$$
P(A) = \sum_i P(B_i \cap A)
$$
The probability of a failure of $i$-th machine depends on the probability of picking this machine $\pi_i$ and the probability of failure $p_i$. The events are independent, so jointly the probability is $\pi_i p_i$, and the overall probability of failure is
$$
p = \sum_i \pi_i p_i
$$
That's the probability we would plug-in into the Geometric distribution to calculate the expected number of runs till failure $1/p$.
Let's verify this using simulation
pi <- c(1/3, 1/3, 1/3)
p <- c(10/100, 20/1000, 50/300)

sim <- function(n) {
    for (i in 1:n) {
        # pick machine
        k <- sample.int(3, size=1, prob=pi)
        # simulate failure or no failure with probability p[k]
        if (runif(1) < p[k]) {
            return(i)
        }
    }
    warning("no success")
}

That gives us the following result that confirms the analytical solution:
> set.seed(42)
> summary(replicate(50000, sim(1000)))
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00    3.00    7.00   10.44   14.00  116.0
> 1/sum(pi * p)
[1] 10.46512

