Question Is this supposed to mean, that I have to actually find Poisson distributed random variable that for a fixed binomial distributed random variable approximates it for the values for which their images coincide (note that a Poisson distributed random variable necesserily has as image $\mathbb{N}$ (with $0$) but a binomial distributed random variable has as image only $\left\{ 0,\ldots,n\right\} $) ?
How is this approximation to be understood (in the illustrative example I have formally written it out) on the level of random variables, since it seems to me, that this is just a "pointwise" (meaning the points in the image of the random variable) approximation...
Illustrative Example I have to calculate the using the binomial and Poisson distribution the probability of getting at most $4$ times the number $1$ in a series of $1000$ games played games, where in each game, randomly on number ist picked ot of the set $\left\{ 1,\ldots,50\right\} $.
I know, that if $X$ is the random variable that counts if how often a $1$ has come up, then the distribution of $X$ is the binomial distrubution, since each instance of a games is a Bernoulli trial (either a $1$ has come up - with probability $p=\frac{1}{50}$ - or it hasn't), so I only have to calculate $ \sum_{k=0}^{4}P\left(X=k\right) $
which is not a difficult task!
But for the Poisson distribution a problem arises: We proved in our course a theorem concerning the binomial and Poisson distribution, that says, that the latter approximates the former, if the number of Bernoulli trials is very big. Formal statement: If $p_{n}$ is a sequence in the interval $\left[0,1\right]$ and $np_{n}\rightarrow\lambda$, then $ P\left(X=k\right)=\binom{n}{k}p_{n}^{k}(1-p_{n})^{n-k}\rightarrow e^{-\lambda}\frac{\lambda^{k}}{k!}\ \text{for}\ n\rightarrow\infty. $
Now I could of course just say that $ P\left(X=k\right)\approx e^{-\lambda}\frac{\lambda^{k}}{k!}\ \text{for}\ \lambda=np $
and then just calculate $ \sum_{k=0}^{4}e^{-np}\frac{(np)^{k}}{k!}. $
But this just doesn't seems right, since there isn't any random variable involved, that actually is $"e^{-np}\frac{(np)^{k}}{k!}"$-distributed. And I thought I would have to exhibit such a random variable and then somehow use the above approximation, because their images have (as noted at the start) different cardinalities.