Probability using Eulers Rule

Question

I have read about Eulers rule but I am still unsure how to do this question ($\frac{1}{n}$ is a fraction) If $X ∼ B(n,\frac{1}{n})$ where $n = 824519$, estimate the probability that $X = 0$ using Euler’s number?

Thank you in advance

Answer 1

When $X\sim B(n,p)$, you have $P(X=k)={n\choose k}p^k(1-p)^{n-k}$.

For $p=1/n$ and $k=0$, you get:

$$p(X=0)=\left(1-\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{} e^{-1}$$

For large $n$, you can thus take the approximation $p(X=0)\simeq e^{-1}$.

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In case you want something more accurate, write

$$P(X=0)=\left(1-\frac{1}{n}\right)^n=\exp\left(n\log(1-\frac1n)\right)=\exp\left[-n\left(\frac1n+\frac1{2n^2}+O(\frac1{n^3})\right)\right]\\=\exp\left(-1-\frac1{2n}+O(\frac1{n^2})\right)=e^{-1}\exp\left(-\frac1{2n}+O(\frac1{n^2})\right)=e^{-1}\left(1-\frac1{2n}+O(\frac1{n^2})\right)$$

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For comparison, for $n=824519$, and to $15$ decimals:

• $P(X=0)=(1-1/n)^n \simeq 0.36787\,92180\,84026$
• $e^{-1}(1-1/2n) \simeq 0.36787\,92180\,84\color{red}{139}$
• $e^{-1}\simeq 0.36787\,9\color{red}{4411\,71442}$