Poisson probability of observing at least one zero out of k independent cases I have a set of observations from a biological experiment. I have counted the occurrences of a  particular event on each of the chromosome for $n$ cells and finally I have the final mean value for each chromosome. 
2.2, 6.4, 3.4, 10.2, 4.45

So I have 5 chromosomes making on an average of above mentioned number of events. 
I have calculated the Poisson probability of any of the above chromosome becoming zero. 
I have used R function: ppois(q=0,lambda=wt[1:5]) and got probabilities 1.108032e-01 1.661557e-03 3.337327e-02 3.717032e-05 1.167857e-02 , which shows there is total probability of 0.15 so to observe any chromosome with zero event.
Now I want to know at what values, these probabilities become zero or negligible. How can I proceed to achieve this?
 A: ## Just repeating what you did for completion
> wt <-  c(2.2,6.4,3.4,10.2,4.45)
> p  <-  ppois(q=0, lambda=wt)
> 
> sum(ppois(q=0,lambda=wt))
[1] 0.1575537
> 
> tol <- .Machine$double.eps # 2.220446e-16
> wt  <- seq(0,250, by=0.01)
> wt[which( ppois(q=0, lambda=wt) < tol )[1]]
[1] 36.05

So when the poisson rate is about 36, the probability that the random variable is 0 is minuscule. Of course you can consider different tolerance levels (I used the smallest number that my computer's R could handle, but that is pretty extreme)
This is a pretty ad hoc way of going about it, and you can probably formalize it with a hypothesis test. 
A: The answer depends on your question.
Here's how I've interpreted it:  
You're got 5 chromosomes, you're interested in the event that any one of them is zero independent of the rest of them, and you're calling this the "total probability".  You'd like to know what combination of individual rates ensures that this total probability is zero or negligible.
The key is what you mean by negligible.  
If you are willing to accept a 1% chance that none of them are zero, then your answer is lambda=6.3, with all chromosomes having a uniform rate.
 > sum(dpois(x=0,lambda=rep(6.3,5)))    # uniform rate for all 5 chromosomes
 [1] 0.009182           # less than 1% total probability

For a 5% chance, the answer is lambda=4.7.
The individual rates, however, could be smaller or larger, provided they contribute to keep the total within tolerance.  For example:
 > sum(dpois(x=0,lambda=c(6,6,7,8,8)))  # non-uniform rates
 [1] 0.00654   # also less than 1% total probability


Solution (uniform rates):
Define a function that does the calculation.  Here I've assumed all rates are equal, though you can set lambda directly to rate and pass in any vector of rates.
 all_chrom_zero <- function(rate) {
     sum(dpois(x=0,lambda=rep(rate,5)))
  }

Then use sapply to run it over a list of trial rates, and choose the smallest rate that comes in under your confidence threshold -- in this case 1%
 tol <- 0.01
 wt<-seq(0,10,0.1)   # candidate rates
 min(wt[which(sapply(wt,all_chrom_zero)<0.01)])
 [1] 6.3


The above code has been structured to take a matrix without too much difficulty (using sapply). So you could do away with the uniform assumption entirely and actually run your code to simulate over various 5-tuples of rates.
This would be a plus if your biological model has information about the distribution of the rates of the individual chromosomes.  If not, you could draw random samples from a Gaussian distribution for each chromosome's rate, each centred at the uniform solution, and run the calculations over these, or if you like brute force calculations, you could (just for fun!) also run the calculation over a 5D grid and look for the "equiprobable" curves at various tolerance levels.
Background for sapply
Using sapply allows one to avoid explicit looping in R and is quite powerful for building simulations.  If you want a bit of background, I've referenced some useful primers in an answer here.
