An urn contains red, green, and blue balls. Let r, g, b be the
proportions of red, green, blue balls, respectively (r + g + b = 1).
A. Balls are drawn randomly with replacement. Find the probability
that the first green ball is drawn before the first blue ball is
drawn. [Hint: Explain how this relates to finding the probability
that a draw is green, given that it is either green or blue.]
For this first question we have with replacement. because the red ball is irrelevant, we can ignore it. so the probability is $\frac{g}{g+b}$.
B. Balls are drawn randomly without replacement. Find the
probability that the first green ball is drawn before the first blue
ball is drawn. Is the answer the same or different than the
answer in (a)? [Hint: Imagine the balls all lined up, in the order
in which they will be drawn. Note that where the red balls are
standing in this line is irrelevant.]
Note that for this question we are drawing the $\textbf{first green ball}$ before the first blue ball. here we also ignore the red balls. $\frac{\binom{g}{1}}{\binom{g+b}{1}}$ which is the same for part a) (???) is this right? if we consider the red balls we can derive $\frac{\binom{g}{1}\binom{r}{k-1}}{\binom{g+b+r}{k}}$.
now part C states:
Generalize the result from (a) to the following setting:
Independent trials are performed, and the outcome of each trial is
classified as being exactly one of type 1, type 2,..., or type n, with
probabilities p 1 , p 2 ,..., p n , respectively.
– Find the probability that the first trial to result in type i comes
before the first trial to result in type j, for i not equal j.
here just consider $t_i$ and $t_j$ for type i, and type j, ignoring the rest. then my work suggests $\frac{t_i}{\binom{t_i+t_j}{1}}$ or alternatively we can derived $\frac{\binom{t_i}{1}\binom{N-t_j}{k-1}}{\binom{N}{k}}$. where N=$t1+t2+...+tn$
For C, relabeling if necessary we may assume that $i=1$ and $j=2$.
Then we can group all of $3,4,\dots,n$ in a single label $l$ -- this is the label we're not interested in.
Do you think you can take it from here? Do you see how this reduces the problem to the first case?