Binomial Distribution Question using Poisson

poisson distributionprobabilityprobability distributionsprobability theory

For this question, I have a general idea of what's going on. The question is: a person typing a document has an error rate of 3 words per 1000 words. If this person typed 6 documents, each with 1000 words (number of errors being independent of each other), what is the probability that 5 documents contain at least 1 error.

First of all, to calculate the probability of 1 document containing at least 1 error, we get calculate the conjugate of $1-$probability of getting $0$ errors. This is equivalent to $\frac{e^{-\mu}\mu^0}{0!}$, where $\mu = \frac{3}{1000}$.

Next, I will take the probability of what we just found and apply it to a Binomial Distribution of Binom($5$, $p$), because this question asks for the probability of 5 documents having at least 1 error.

I wanted to ask if my findings and calculations are on the right path. Any help would be appreciated ^.^

Best Answer

I think you're on the right track, but the Poisson parameter is 3...not 3/1000.

On a given page the probability of at least one error is

$1-\frac{e^{-3}3^0}{0!}=1-0.0498=0.9502$, so the probability that $5$ documents have at least $1$ error (and presumably the 6th page has no errors) is

$\binom{6}{5} \; 0.9502^5 \; (0.0498)^1 = 0.2314$

Related Solutions

Using the Poisson approximation to calculate probability

One question is whether 'has 5 errors' means 'exactly 5' or 'at least 5'.

Poisson model. As you say, if $X \sim \mathsf{Pois}(\lambda = 3),$ then $P(X = 5) = 0.1008.$ But $P(X \ge 5) = 1 - P(X \le 4) = 0.1847.$ [Computations in R, where dpois is a Poisson PDF and ppois is a Poisson CDF.]

dpois(5, 3);  1 - ppois(4, 3)
## 0.1008188
## 0.1847368

Possible normal approximation. For large or moderate $\lambda,$ the distribution $\mathsf{Norm}(\mu = \lambda,\, \sigma=\sqrt{\lambda})$ might be used as an approximation to $\mathsf{Pois}(\lambda),$ but $\lambda = 3$ is a little too small. If $X^\prime \sim \mathsf{Norm}(3, \sqrt{3}),$ then $P(X^\prime > 4.5) = 0.1932,$ which is not a horrible approximation of $P(X \ge 5) = 0.1957.$

Since you are reviewing for an exam, you might practice the normal approximation for larger values of $\lambda,$ perhaps using standardization and printed normal tables.

The figure shows PDFs of both the Poisson distribution and (roughly) approximating normal distribution for the question at hand.

1 - pnorm(4.5, 3, sqrt(3))
0.1932381

Speculative binomial model. I do nut understand why the Binomial distribution is mentioned. Under some circumstances, a binomial probability can be approximated by using a Poisson distribution with $\lambda = np,$ but I don't see a value of $n$ in the statement of your Question.

If somehow $Y \sim \mathsf{Binom}(n = 1000, p = 0.003),$ then $P(Y \ge 5) = 0.1845 \approx 0.1957$ (from the Poisson distribution).

This might be a reasonable model if you imagine the document has $n = 1000$ opportunities for errors, each with probability $p = 0.003,$ giving the binomial mean $\mu = np = 3.$

In the figure, the open blue dots show the relevant binomial probabilities. [At the resolution of the figure, about 2 decimal places, it is difficult to distinguish between the Poisson and binomial models.]

1 - pbinom(4, 1000, 0.003)
## 0.184484

Best Answer

Related Solutions

Using the Poisson approximation to calculate probability

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