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Suppose that, on average, a bushel of 100 apples contain three apples with a worm. Select a bushel of 100 apples at random. Let $Worm$ be the random variable that represents the number of apples with a worm. Assume that the number of apples that contain a worm can be modeled by a binomial random variable with parameters n=100, and p=0.03. Find $\Pr(Worm=k)$ for $k=0,1,2,3,4,8$. Compare these with the Poisson model.

For the binomial model, I know that the equation is as follows:

$\Pr(X=k)=\binom{100}{k}(0.03)^k(0.97)^{100-k}$

I was having a little trouble developing the equation with the Poisson model, specifically the lambda parameter. I understand that lambda is the rate of occurrence. For example, we might know that there are 500 misprints in a 250 page manuscript, which translates into an average rate of $\lambda=500/250=2$ misprints per page.

In this problem, I thought $\lambda=3/100$ giving the following equation:

$\Pr(Worm=k)=e^{-(3/100)} \cfrac{(-3/100)^k}{k!}$

but apparently $\lambda = 3 $. Can someone please explain why? I thought lambda is a rate so it should be 3 rotten apples per a bushel of 100 apples, not (3/100)*100 = 3....

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The Poisson with paramteter $\lambda$ has mean $\lambda$. So when we are approximating a binomial with parameters $n$, $p$ (and therefore mean $np$) by a Poisson, the appropriate parameter $\lambda$ is the mean $np$ of the binomial. But the above can mainly be thought of as a mnemonic, a device to remember the right answer. So we go into more detail.

The following is an informal calculation that can be turned into a formal limit argument. If $X$ is the binomial, then $$\Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}.\tag{1}$$ Use the abbreviation $\lambda$ for $np$. Then $p=\frac{\lambda}{n}$. So we have $1-p=1-\frac{\lambda}{n}$.

We can then rewrite (1) as $$\Pr(X=k)=\frac{1}{k!}(n)(n-1)\cdots(n-k+1)\frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^n \left(1-\frac{\lambda}{n}\right)^{-k}.\tag{2}$$ Note that we have done nothing so far, except for the introduction of $\lambda$ as an abbreviation for $np$.

Now imagine $n$ large, and $p$ small, so that $np=\lambda$ stays constant. Let $k$ be fixed. We look at the various terms in Formula (2).

For fixed $k$, if $n$ is large then $(n)(n-1)\cdots(n-k+1)\frac{1}{n^k}\approx 1$ and $\left(1-\frac{\lambda}{n}\right)^{-k}\approx 1$. Also, for $n$ large, $\left(1-\frac{\lambda}{n}\right)^n \approx e^{-\lambda}$.

It follows that the right-hand side of (2) is approximately $$\frac{1}{k!} \lambda^ke^{-\lambda}.$$ This is precisely the probability that a Poisson with parameter $\lambda$ takes on the value $k$.

In your example, the mean number of rotten apples in a batch of $100$ is $3$, and $n=100$ is reasonably large, so one can expect the Poisson with parameter $3$ to provide a good fit, at least for reasonable $k$.

Remark: We gave a fair bit of detail justifying why in the case $n$ large, $p$ small, and $np=\lambda$ moderate, we can approximate binomial distribution probabilities by Poisson distribution probabilities. But the main answer to your question is that the appropriate $\lambda$ is always the mean of your binomial. If a superbushel holds $150$ apples, the appropriate $\lambda$ is $4.5$.

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