The probability mass function of X

probability

I'm having trouble with the following question: Initially we have 1 red and 1 green ball in a box. At each step we choose a ball randomly from the box, look at its color and then in put it back into the box with an additional ball of the same color. This way there will be 3 balls in the box after the first step, 4 balls after the second step, and so on. Let X denote the number of red balls in the box after the 1000th step. Find the probability mass function of X.

Here's where I started: it's possible that the box can end up with one red ball, in which case we would have chosen 999 green balls, two red balls, in which case we would have chosen one red ball and chosen 998 green balls, … etc.

For the first case, for example, I took the probability of the box ending up with only one red ball as the probability of choosing 999 green balls. I just want to know if I'm on the right track here… Any help would be appreciated.

Best Answer

It can be proved by means of induction that after $n$ steps the number of red balls in the box have uniform distribution on $\{1,\dots,n+1\}$.

Observe that under assumption that this is the case for a fixed $n$ then

  • the probability on $1$ red ball after a next step equals: $$\frac1{n+1}\frac{n+1}{n+2}=\frac1{n+2}$$
  • the probability on $n+2$ red balls after a next step equals: $$\frac1{n+1}\frac{n+1}{n+2}=\frac1{n+2}$$
  • the probability on $r\in\{2,\dots,n+1\}$ red ball after a next step equals: $$\frac1{n+1}\frac{r}{n+2}+\frac1{n+1}\frac{n+1-r}{n+2}=\frac1{n+2}$$

This illustrates that $P(n)\implies P(n+1)$ where $P(n)$ is the statement that must be proved.

The base case is easy.

So $X$ has unifom distribution on $\{1,\dots,1001\}$.

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