Poisson Distribution

I've a question here that I may possibly help with:

"Suppose that the number X of errors in an assignment submitted by each student in a certain group is a random variable that has a Poisson Distrubition with parameter Lambda = 4. Suppose also that all the assignments submitted are independent of one another.

(a) What is the probability that a given submitted assignment contains no errors?

(b) What is the probability that, among 10 assignments submitted, there are at least two that contain no errors?

Now, our lecturer gave us the answer which is:

Answer: Let Y = Number of assignments among the 10 submitted that contain no errors. Then Y has a binomial distribution with parameters n = 10 and p = P[X=0].

However, I would like to elaborate on this answer. Is this approach correct?

Answer for (a): Let k = 0 and Lambda = 4 in our Poisson formula, yielding e^-4 (which I think is the probability that a given assignment contains no errors).

for (b): Use the Binomial pmf (in the context mentioned above), with n=10, p=e^-4, 1-p = 1-e^-4 and compute the probabilities for k = 0 and 1 respectively. (i.e. No assignments are free of errors, only one assignment is free of errors) I obtained P(k=0) as 0.831225... and P(k=1) as 0.155085... Adding these two together should yield 0.98631.., now subtract this from one to find the probability of at least two assignments with no errors as 0.01369.

Thank you. Just to clarify, this is not homework; it's just important I get the method right.

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yes, this seems correct. There is no better approach than this. –  Ronald Apr 21 '12 at 19:40

1 Answer

Yes, you got it right.

There are 10 students. Let $X_i$ be the random number of errors in assignment submitted by $i$-th student. The problem states that $X_i$ are independent identically distribution Poisson random variables with mean $\lambda = 4$. $$\mathbb{P}(X_i = k) = \mathrm{e}^{-\lambda} \frac{\lambda^k}{k!}$$ Let $U_i$ be indicator variables, equal to 1 is $i$-th assignment is error-free, and zero otherwise, i.e. $U_i = I(X_i=0)$. $U_i$ are independent identically distributed random variables, with $$\mathbb{P}(U_i=0) = \mathbb{P}(X_i=0) = \mathrm{e}^{-4} \quad \mathbb{P}(U_i=1) = \mathbb{P}(X_i > 0) = 1- \mathrm{e}^{-4}$$ that is $U_i$ are iid Bernoulli random variables with $p=\mathrm{e}^{-4}$.

The number of errors in $n=10$ submitted assigments follows Binomial distribution $$Y = \sum_{i=1}^{n} U_i \sim Bi(n,p) \qquad \mathbb{P}(Y =k) = \binom{n}{k} p^k (1-p)^{n-k}$$ Therefore $$\mathbb{P}(Y \geqslant 2) = 1 - \mathbb{P}(Y=0) - \mathbb{P}(Y=1) = 1-(1-p)^n - n p(1-p)^{n-1}$$

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