Balls in bins – approximation of Binomial distribution by Poisson distribution

balls-in-binsbinomial distributionpoisson distributionprobability

In our lecture we have the following example to illustrate the approximation of a Binomial distribution by Poisson distribution.

We have $10000$ bins which contain black balls. We know that $5000$ red balls are distributed among the $10000$ bins. We choose a bin and ask for the probability that it contains at least one red ball. The probability that a red ball is in a particular bin is $p=\frac{1}{10000}$.

My approach:

I can also interpret the example such that I randomly distribute $5000$ numbered balls one after another among the $10000$ bins. My intution tells me that the probability should be very high that at least one ball is placed in the chosen bin.

Indeed,

if I define $\Omega:=\{1,\dots,10000\}^{5000}$, then $|\Omega|=5000^{10000}$, where each $\omega\in\Omega$ has the same probability $p(\omega)=\frac{1}{10000^{5000}}$. We immediately see that $5000^{9999}$ are all possibilites where the selected bin in the beginning doesn't contain one of the $5000$ balls. Hence, the probability we are looking for is simply $\frac{5000^{10000}-5000^{9999}}{5000^{10000}}=1-\frac{1}{5000}\approx 1$.


However, the professor defined a random variable $X_k$ which attains $1$ if the $k$-th red ball ends up in our chosen bin or $0$ if not. So she wrote
$$
P\left(\sum\limits_{k=1}^{5000}X_k=0\right)\approx e^{-5000\cdot\frac{1}{1000}}=e^{-0.5},
$$

where $\approx$ indicates that the Poisson approximation was used. So in her description the probability that the chosen bin contains at least one red ball is $\approx 1-e^{0.5}$ which is signifcantly smaller than the probability I calculated.


Where is my mistake?

Best Answer

You have probably mixed up the base and the exponent in some of the values.

For example, $|\Omega|=10000^{5000}$ because every ball has 10,000 choices and there are 5,000 balls.

Then, say bin #1 is the chosen bin. That no balls are in bin #1 means that all balls go to the remaining 9,999 bins. The number of possibilities is $9,999^{5000}$.

Hence, the required probability is

$\dfrac{10000^{5000}-9999^{5000}}{10000^{5000}}=1-.9999^{5000}\approx\underline{\underline{.606515}}$

Compared to $e^{-.5}\approx.606531$, they are very close to each other.

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