Binomial -> Poisson distribution question

[sunrah]

Homework Statement

A teacher has an infinite flow of papers to mark. They appear in his office at random times, at an average rate of 10 a day. On average 10% of the manuscripts are free from errors. What is the probability that the teacher will see exactly one error-free manuscript (a) after he has marked 10 of them? (b) after a day?

Homework Equations

Binomial dist.:
$P(k) = \frac{N!}{k!(N-k)!}p^{k}(1-p)^{N-k}$

Poisson dist.:
$P(k) = \frac{\mu^{k}}{k!}e^{-\mu}$

The Attempt at a Solution

a)
The first part seemed straight forward, just plug on following values
No. of trials: N = 10
No. of successes: k = 1
Prob. of success: p = 0.1

This gives an answer ~0.4 (1 s.f.)

b) The second part I don't really understand. I know it goes to a Poisson distribution because I asked my teacher if that is so. I thought P-distributions are for very large N? Can we assume that here? Mean I know the no. of papers i infinite, but the per day average is only 10?

Secondly is the mean value per day μ, the same as the expectation value per 10 papers? I don't think so, but here goes:
let μ = 0.1 and using second equation -> P(1) = ~0.2 (1 s.f)

I thought given the same parameters, the Poisson and Binomial distributions should give very similar results.

 

[DrDu]

No, P-distributions are not restricted in any sense to very large N.

 

[sunrah]

No, P-distributions are not restricted in any sense to very large N.

Sorry, I meant that the Poisson is the limit of the binomial distribution for large N, but here N = 10?

 

[Ray Vickson]

Sorry, I meant that the Poisson is the limit of the binomial distribution for large N, but here N = 10?

No, N is not 10, it is ∞. The question said the teacher marks an infinite flow of papers.

However, I think the problem leaves out some important and significant information: it says the papers appear at random times and at a given average rate. We are likely being asked to assume that "random times" mean "independent, identically distributed inter-arrival times", but even then we do not really have a rigorous justification of the Poisson arrival model.

Nevertheless, I would bet the person setting the problem wants us to assume Poisson arrivals, despite not having enough information to justify it. I think that is acceptable in an introductory course.

 

[DrDu]

Sorry, I meant that the Poisson is the limit of the binomial distribution for large N, but here N = 10?

That you can derive the Poisson distribution as a limit of some other distribution doesn't say nothing about it's range of validity.

 

[DrDu]

Sorry, I meant that the Poisson is the limit of the binomial distribution for large N, but here N = 10?

That you can derive the Poisson distribution as a limit of some other distribution doesn't say nothing about it's range of validity.

 

[sunrah]

Nevertheless, I would bet the person setting the problem wants us to assume Poisson arrivals, despite not having enough information to justify it.

Yes, that's correct. I'm still none the wiser about part 2. If we assume Poisson distributed data, it still needs a mean μ over one day and this what I don't get. It seems like μ = Np = 0.1, but over the day the number of papers N that arrive may be less or greater than 10??

 

[RUber]

I agree with your part (a).
For part (b) It looks like you will have an infinite sum. k = 1, and P(k) will still follow the binomial distribution.
However, you need to handle N.
N is Poisson distributed.
Total probability $P = \sum_{N=0}^\infty P_{poisson}(N) *P_{binom}(k|N)$
This sum will necessarily converge. It may take you some additional work to find it. Otherwise, just plug it into Excel and see what happens when you start getting N up near 20 or so.

 

[Ray Vickson]

Yes, that's correct. I'm still none the wiser about part 2. If we assume Poisson distributed data, it still needs a mean μ over one day and this what I don't get. It seems like μ = Np = 0.1, but over the day the number of papers N that arrive may be less or greater than 10??

The average arrival rate is 10 per day = 10/24 per hour = 10/(24*60) per minute. So, in a day of 24*60 = 1440 minutes, we have an arrival probability of p = 10/1440 = 0.006944 in each minute. This is a case of a binomial with a large N (N = 1440) and a small p (p = 10/1440), so leads right away to the Poisson limit.

And, of course, the number of arrivals in any particular day can be any integer from 0 to ∞, but will very likely not differ too much from 10.

 

[sunrah]

thank you that makes sense. so if we're work out the probability of an error free paper per minute, then each trial can be a time interval of 1 min and still with two possible outcomes - success or failure.

 

[Ray Vickson]

thank you that makes sense. so if we're work out the probability of an error free paper per minute, then each trial can be a time interval of 1 min and still with two possible outcomes - success or failure.

Warning: the analysis I gave you was for paper arrivals of all types---whether or not they have any errors.

If you want to look at only papers with no errors, you need to use a different rate and a different Poisson process. As before, assuming the error/non-error probabilities are independent of everything else, the average number of error-free papers is (1/10)*(1) = 1 per day = 1/24 per hour = 1/(24*60) = 1/1440 per minute = 1/(24*60*60) = 1/86,400 per second, etc. So, in a day of N = 86,400 seconds, with an error-free arrival probability of 1/86,400 each second (and assuming independence) we are in the very large-N, very small-p limit, giving a Poisson random variable $X$ with mean $\lambda = 86,400/86,400 = 1$ (per day). So, the probability of exactly 1 error-free paper in a day is $P(X = 1)$, where $X\sim \: \text{Poisson}(\lambda = 1)$.