Expected no of balls to select before a certain type of ball comes

Question

There are w white balls and r red balls in a box, to find the expected no of balls to pick before we get a red ball? $$\qquad$$ What I have tried is, Let $ X_k $ denote that k no of white balls have been taken out before we get a red ball, clearly k= {0,1,2...,w} , now if X denotes the no of balls we have to pick before we get a red ball, then we have by linearity of expectation, E(X) = $ E(X_1) + E(X_2) +...+E(X_w) $ , now for $ X_k $ we have $$ E (X_k) = k {\frac { {\binom {w} {k}} * {\binom {r} {0}} } { \binom {w+r} {k} }} \frac {r}{r+w-k} $$ I couldn't get a simple answer on how to simplify the sum of this expression over k=0 to w, any help on what to do next and what i have tried is right?

Answer 1

Imagine taking all the balls out in some random order and lining them up in the order they appear. The $r$ red balls act as separators for $r+1$ 'runs' of white balls (where a 'run' may be empty, contrary to the usual meaning of a run). Thus $w$ white balls are distributed among these $r+1$ runs. So the average run of white balls is of length $\frac{w}{r+1}$.

To find the first red ball, we must go through the first run of white balls, and so the average (expected) position of the first red ball is $\frac{w}{r+1}+1$.

Answer 2

We need to take a weighted average of events. That is, $\frac{r}{r+w}$ percent of the time, we will only have to draw one ball. Then $\frac{r^2}{(r+w-1)(r+w)}$ percent of the time it takes 2. This goes on and on and we need the weighted average, which, in closed form is$$ \frac{r}{w}\sum_{i=1}^{w}\frac{iw^{i-1}}{\prod_{k=0}^{i-1}(r+w-k)}=\frac{r}{w}\sum_{i=1}^{w}\frac{iw^{i-1}(r+w-i)!}{(r+w)!}$$

Answer 3

The probability that any one white ball gets picked is $1/(r+1)$.
The average number of white balls that get picked is $w/(r+1)$.