Which distribution to use for a probability problem? 
In a food processing plant, on average, 2 packaging machines breakdown per week.

I let x be the number of processing machine breakdowns in a week. So x is Poisson(2), but one part of the question says: 

in a random sample of 26 machines what is the probability of no machine breakdowns in at least one of them in a week.

I know, mean = 2 and solve 2=26*p to get p=1/13. Then I say x is Binomial(26, 1/13). 
But now I do not know where to go from there. Any suggestions please? I am not interested in the final answer, I just need to be pointed on the right track from where I am. 
 A: The following analysis illustrates one approach to obtaining a solution.  At least it might help show how to work with the Poisson distribution.
To answer this question constructively and clearly, let's make a few simplifying assumptions to avoid getting bogged down in details that haven't been described.  For instance, you might choose to


*

*assume that a "breakdown" is an event with such a short duration that a machine is back in operation immediately after a breakdown; and

*therefore the same machine could break down multiple times during a week (although this might be a rare event).
As apparently intended by the question, we will make some additional stronger assumptions.  Some such (modeling) assumptions are needed to make any progress at all with the answer.  Their chief purpose would be to give us a point of departure for eliciting additional information from the plant engineers so we could develop improved models and better answers:


*

*All machines independently have the same chances of breaking down and

*those chances do not vary over time.
These assumptions imply the number of breakdowns observed among any number $N$ of machines during any period of $x$ weeks has a Poisson$(\lambda N x)$ distribution, where $\lambda$ is a parameter common to all machines at all times.  The question tells us about the breakdown rate for $x = 1$ week:
$$\lambda N 1\text{ weeks} = 2.$$
Therefore 
$$\lambda = 2 / (N \text{ machine-weeks}).$$
In a random sample of $26$ such machines, the number of breakdowns in a week will have a Poisson distribution with parameter
$$\mu = \lambda\times (26\text{ machines})\times (1\text{ week}) = 26\lambda = 52/N.$$
From the formula for Poisson probabilities, the chance of no breakdowns among these $26$ machines is
$$e^{-\mu} 0! = e^{-\mu} = e^{-26\lambda} = e^{-52/N}.$$
Since $N\ge 26$, this value cannot exceed $e^{-52/26}=e^{-2}\approx 0.135$, but as $N$ grows large it could become arbitrarily small.
This is not a final answer. It only shows the implication of four assumptions that were made upon interpreting the question in terms of the chance of no breakdowns in a week.  (Other interpretations of the question are possible, due to the contorted syntax used to pose it.)  In particular, the dependence upon the unknown total number of machines is clear and explicit.  This is about as far as one can go, given the limited information supplied in the question.

A simulation (covering almost 200 years of operation) illustrates the ideas.  Its output consists of two histograms: the weekly breakdown counts for all $N$ machines and the counts for the sample of the machines.  Here is an example for $N=60$:

On each histogram are drawn two vertical lines: a gray one indicating the location of the actual rate (as given by the preceding solution) and a dashed red one indicating the average rate during the simulation.  In each case those lines are visibly coincident, showing that the simulation and the preceding analysis are in agreement.
Studying the R code that produced this simulation may help clarify the ideas.
n <- 60           # Number of machines
sample.size <- 26 # Must be less than or equal to n
weekly.mean <- 2  # Events per week, on average
n.iter <- 1e4     # Size of this simulation in weeks
set.seed(17)      # Reproduce these results exactly
#
# Simulate all machines.
#
lambda <- weekly.mean/n                          # Weekly breakdown rate per machine
x <- matrix(rpois(n.iter*n, lambda), nrow=n)     # Breakdowns by machine by week
weekly.breakdowns <- colSums(x)                  # Total breakdowns each week
sample.breakdowns <- colSums(x[1:sample.size, ]) # Total breakdowns in the sample
#
# Plot the results.
#
par(mfrow=c(1,2))
eps <- 0.99
hist(weekly.breakdowns, breaks=(-1):max(weekly.breakdowns)+eps,
     freq=FALSE, cex.main=0.9)
abline(v=lambda * n, lwd=2, col="Gray")
abline(v=mean(weekly.breakdowns), col="Red", lwd=3, lty=3)

mu <- weekly.mean * sample.size / n
hist(sample.breakdowns, breaks=(-1):max(sample.breakdowns)+eps,
     freq=FALSE, cex.main=0.9)
abline(v=mu * n, lwd=2, col="Gray")
abline(v=mean(sample.breakdowns), col="Red", lwd=3, lty=3)

A: Both the Poisson and the binomial distribution can model counts, but they do so differently.  You assume that each machine can break down or not at most once each week.  That suggests a particular distribution is more appropriate.  
From this, and your calculations, you believe the right distribution is binom(26, 1/13).  You should be able to calculate probabilities directly from that distribution, now that you have settled on one.  
The probability you are asked to calculate is somewhat ambiguous.  The phrasing in the body of the question ("the probability of no machine breakdowns") is inconsistent with what you report in the comments ("the probability that at least one machine doesn't breakdown").  These two would be $P(X=0|{\rm binom}(26, 1/13))$ and $P(X\le25|{\rm binom}(26, 1/13)$, respectively, which are not equal.  Irregardless of which is what you were asked, both should be straightforward to calculate from the distribution you have decided on.  
