Coin toss Poisson approximation

Question

Let's say we look at independent $1000$ fair $0$/$1$ coin tosses of the length $10$. Now we want the probability that exactly $2$ of these $1000$ $0$/$1$ sequences are consisting only of $1$'s. I have to use Poisson approximation.

$$\mathbb{P}(Y=k)\approx \frac{\lambda^k}{k!}e^{-\lambda}$$

and for $\lambda$ we use the Binomial distribution $\lambda=n\cdot p$

$$\lambda=1000\cdot \underbrace{0.5^{10}}_{probability \ of \ only \ getting \ 1's}=\frac{125}{128}$$

$$\mathbb{P}(Y=2)\approx \frac{\left(\frac{125}{128}\right)^2}{2!}e^{-\left(\frac{125}{128}\right)}=0.1795$$

Will be this approach correct for this task? Will be this the correct Poisson approximation?

Answer 1

Yes - this is the desired approach using a Poisson approximation.

As a check using R, the Possion approximation gives

> dpois(2, lambda=1000*2^-10)
[1] 0.1795785

while the binomial calculation would be

> dbinom(2, size=1000, prob=2^-10)
[1] 0.1796641