1
$\begingroup$

An urn contains one black and one white ball. Someone draws a ball and then places it back and adds as many white balls as are currently in the urn. This is process is repeated infinitely often.

What is the probability that there is a stage at which no more black balls are drawn. I was given the tip of using the Borel-Cantelli Lemma.

My idea:

Define random variable at each time $i$, namely $X_{i}=1$ if the ball drawn is white, while $X_{i}=0$ if the ball drawn is black.

Note that the sequence of random variables $(X_{i})_{i \in \mathbb N}$ is independent. Then I get stuck because looking at either sum:

$\sum_{i \in \mathbb N}P(X_{i}=1)=\sum_{i \in \mathbb N}\frac{2i-1}{2i}=\infty$

$\sum_{i \in \mathbb N}P(X_{i}=0)=\sum_{i \in \mathbb N}\frac{1}{2i}=\infty$

and neither helps me in drawing the wanted conclusion. Any ideas?

$\endgroup$
2
$\begingroup$

$X_i$ is the probability to draws a black ball after the first draw, one ball is added so there are $2$- white balls, after the second draw, $2$-balls are added there are $4=2^2$ white balls,... after the $i$ draws, $2^{i-2}$ balls are added to the existing $2^{i-2}$ balls so there are $2^{i-1}$ white balls.

after $i$-draws there are $2^i$ white balls so $X_i={1\over{2^{i-1}+1}}$ you can apply Borel Cantelli since $\sum_{i\geq 1}{1\over{2^{i-1}+1}}$ converges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.