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I tried to solve the following question:

A machine produces Coca Colas. The count of Coca Colas per day is not known in advance but It can be described with a Poisson distribution and a rate of 5 Coca Colas produced daily by the machine. When it produces more than 10 a day it stops working. What’s the probability that it stops at least twice during one week?

With this formula:

$$P(x) = e^{-\lambda*t}\frac{(\lambda*t)^x}{x!}$$

I thought:

The probability of at least two is the probability of receiving two, or three, or four, etc.

$$P(X\geq2)=P(X=2)+P(X=3)+\dots=\sum_{x=2}^\infty P(X=x),$$

the complementary event of "at least 2", which is "at most one":

$$P(X\geq2)=1-P(X<2)=1-P(X\leq1)=1-[P(X=0)+P(X=1)].$$

Then I calculated:

$$P(0) = e^{-5*7}\frac{(5*7)^0}{0!}= $$

$$P(1) = e^{-5*7}\frac{(5*7)^1}{1!}=$$

$$1-[P(X=0)+P(X=1)] = 1 - 0 = 1$$

The result of 1 is wrong. How would you calculate it? Thanks

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  • $\begingroup$ As zoli noted in the answer, the probability you are looking for is not of how many coca-colas the machine produces, but how many times it stops, and it stops after producing 10 or more. So you have to take this into account. $\endgroup$ – Anna SdTC Jan 11 '18 at 0:45
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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

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

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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.

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  • $\begingroup$ Maybe you were supposed to use a normal approximation for $p$ (not a particularly good idea for $\lambda$ as small as 5). With a normal approximation, I got $p = 0.007.$ $\endgroup$ – BruceET Jan 10 '18 at 2:40

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