We have 60 green balls and 40 white balls in an urn.
Let $X_n$ be the number of balls drawn on step n defined as $ X_1 = 10 $ and $X_n =$ the number of green balls drawn in step $n-1$. The balls are then returned to the urn.
I need to calculate $\mathbb{E}(X_n)$ and show that $\lim_{n \to +\infty} \mathbb{E}(X_n) = 0 $. Suggestion: Find a suitable r.v. to condition $X_n$.
Intuitively it makes sense to me that the limit would be 0 since with high probability we're drawing each balls each time but I can't seem to find anything not convoluted for $\mathbb{E}(X_n)$.
Here's my work:
Let $Y_n$ be the amount of green balls drawn in step n. So we have that $X_n = Y_{n-1}$. $X_2 = Y_1 \sim Bi(10,0.6)$ so $\mathbb{E}(X_2) = 6$.
In general $\mathbb{E}(X_n) = \sum_{k=0} ^{10} k \frac{6}{10} P(Y_{n-2} = k)$
But I'm not sure how to go forward since each step depends on all the steps before and it gets very convoluted.
Any help would be greatly appreciated.
The idea is the same as in this other answer. Instead of drawing only as many green balls as were drawn previously, perform the following operation ten times: draw until you get a white ball. Given this scheme, the probability that you get to step $n$, where you then draw a green ball, is $(3/5)^n$, so the expected number of green balls drawn in step $n$ of the original formulation is $10(3/5)^n$, i.e. $E(X_n)=10(3/5)^{n-1}$. It is clear that this goes to zero as $n\to\infty$.