Poisson probability per week

Let $p$ denote the probability that the machine stops at a certain day. Furthermore, assume that the daily workings of the process are statistically independent. Then the probability that the machine stops at least twice a week is $$1-{7\choose 0}p^0(1-p)^7-{7\choose 1}p(1-p)^6=1-(1-p)^7-7p(1-p)^6.$$

Now, are the daily productions of the machine independent? Yes.

It remains to calculate $p$.

The daily production is of Poisson with parameter $\lambda=5$. That is, the probability that the machime produces $k$ coca colas at a day is $e^{-5}\frac{5^k}{k!}.$

The probability that $k\geq 10$ equals one minus the probability that the number of coca colas produced at a certain day is less than $10$:

$$p=1-e^{-5}\sum_{k=0}^9\frac{5^k}{k!}\approx0.032.$$.

So, the probability sought for equals

$$1-(1-p)^7-7p(1-p)^6\approx0.019.$$

Comment. I will work the problem using R statistical software, with a method similar to that of @zoli (+1), who worked a similar problem with machine stoppage after 'ten or more' rather than 'more than ten' per day. You should re-compute the answers I got from R by using the appropriate formulas for Poisson and binomial PDFs in your textbook or notes.

Number of Cokes in a day is $X \sim \mathsf{Pois}(\lambda = 5).$ So the probability $p$ the machine stops on a given day is $p = P(X > 10) = 1 - P(X \le 10) = 0.0137.$

 p = 1 - ppois(10, 5);  p
 ## 0.01369527

Number of times it stops working in a week is $Y \sim \mathsf{Binom}(n = 7, p).$ So you seek $P(Y \ge 2) = 1 - P(Y \le 1) = 0.0038.$

1 - pbinom(1, 7, p) 
## 0.003762613

Note: A Poisson random variable with rate $5(7) = 35$ would be correct for the number of Cokes the machine makes per week, if it didn't stop whenever it made more than 10 Cokes in a day.