An urn contains ${b \ge 1}$ blue balls and one red ball. At each stage a ball is randomly chosen from the urn and then replaced along with a new ball of the same color. Let $T$ denote the first step when a red ball is chosen. What is ${P[T \ge i]}$ and what is $E[T]$?
I calculated $1- P[T < i]$ to get $\frac{b}{b+i-2}$. However, this doesn't allow for me to calculate the infinite sum that is required for $E[T]$ which by my current answer would be $b(\frac{1}{b-1}+\frac{1}{b}+\frac{1}{b+1}+..)$
Would appreciate some help. Thanks!
For $T \ge i$ you must draw a blue ball for each of the first $i-1$ steps.
The probability of this occuring is $$\frac{b}{b+1}\times\frac{b+1}{b+2}\times\frac{b+2}{b+3}\ \times\ ... \ \times\frac{b+i-2}{b+i-1}$$ Hence, $P[T \ge i]$ is equal to $\frac{b}{b+i-1}$
The expected value $E[T]$ is given by $$(1\times\frac{1}{b+1})+(2\times\frac{b}{b+1}\frac{1}{b+2})+(3\times\frac{b}{b+1}\frac{b+1}{b+2}\frac{1}{b+3})+\ ... \ =$$ $$b\sum_{i=1}^{\infty} \frac{i}{(b+i-1)(b+i)}$$