Probability that the red fish are the first species to become extinct I have a doubt in the solution of the next problem:

A pond contains $3$ distinct species of ﬁsh, which we will call the Red, Blue, and Green
  ﬁsh. There are r Red, b Blue, and g Green ﬁsh. Suppose that the ﬁsh are removed
  from the pond in a random order. (That is, each selection is equally likely to be any
  of the remaining ﬁsh.) What is the probability that the Red ﬁsh are the ﬁrst species
  to become extinct from the pond? 

(HINT: Write P{R} = P{RBG} + P{RGB}, and compute the probabilities on the right by ﬁrst conditioning on the last species to be removed. The thing is that P{RBG}=P{RBG|G is last}P{G is last} (the probability that the green fish are the last to become extinct) 
Then the probability that G is last is $${g\over r+b+g}$$
but the probability that the blue fish is the  second species to go extinct given that the green fish is the last species to go extinct P{RBG|G is last} I don´t quite get it, can I ignore the green fish and consider the probability that the blue fish are last amongst the red and blue fish only?
Hence the probability of P{RBG|G is last} would be $$b\over r+b$$
or I can´t ignore the green fish?
I would really appreciate your help.
 A: It may help to "run time backward".  Instead of fish, think of a deck of cards with $R,B,G$ red, blue and green cards, shuffled randomly.  The colour of the $k$'th card in the deck is the colour of the $k$'th fish to be caught.  Now turn the deck over: it's still shuffled randomly, but now the $k$'th card tells you the colour of the $k$'th last fish to be caught.  Turn over the first card: with probability $G/(R+B+G)$ it's green, indicating that the last surviving fish is green.  
Given that the first card is green, which of red and blue will appear next in the deck (indicating the fish colour that is not the first to go extinct)?  Since we don't care about green cards any more, remove all the green cards from the remaining deck, and note that the red and blue cards are still randomly shuffled...
A: Answer:
Consider an fish tank with 2 black fishes and 1 white fish.  We need to find the probability that white fish is the last to become extinct.  A simple enumeration would be 
$B_1B_2W_1$
$B_1W_1B_2$
$W_1B_1B_2$
$B_2B_1W_1$
$B_2W_1B_1$
$W_1B_2B_1$
Of these for white to be the last to become extinct is equivalent to white fish being the last fish to be removed.  In the above enumeration,of the six possible outcomes, there are two outcomes where white is the last.  The probability that White is the last species to be removed is $\frac{1.2!}{3!} = \frac{1}{3}$.  Extending the same for n black fishes and m white fishes,  the total number of outcomes = (n+m)!.  Number of ways the last fish is white is m.(m+n-1)!.  Thus the probability is $\frac{m.(m+n-1)!}{(m+n)!} = \frac{m}{m+n}$.
In your problem, if there are r red fishes, b blue fishes, and g green fishes,
Based on the same reasoning as (a), the probability that the Blue fish and the Green
fish are the last species to become extinct is $\frac{b}{r+b+g}\text{ and }\frac{g}{r+b+g}$ respectively. Furthermore, the conditional probability that the Blue fish are the second species to become extinct, given that Green fish is are the last species in the pond, is$ \frac{b}{r+b}$. Similarly, the conditional probability that the Green fish are the second species to become extinct, given that the Blue fish are the last remaining fish in the pond, is $\frac{g}{r+g}$
Let
R
be the event that the Red fish are the first species to become extinct in the pond,
RBG denote the event that the Red fish first become extinct, followed by the Blue fish, then by the Green fish, and RGB denote the event that the Red fish first become extinct, followed  by the Green fish, then by the Blue fish. Then we can obtain
$P(RBG) = P(G-last)P(RBG/G-last) =\frac{g}{r+b+g}\frac{b}{r+b}$
$P(RGB) = P(B-last)P(RGB/B-last) =\frac{b}{r+b+g}\frac{g}{r+g}$
Finally, we can compute the probability that the Red fsh are the first species to become
extinct in the pond by adding these two.

A: Writing, $RBG$ to mean "red fish go extinct first, followed by blue fish, and finally green fish", and so forth, and using $\circ$ to mean "fish of another colour".
Hang the fish on a string in the order they are caught.  Each fish has the same chance of being the last fish caught.  So the probability that the last fish caught is green is:
$$\Pr(\circ \circ G) = \frac{g}{r+b+g}$$
Now, travel back along the string until you find the last fish that was not green.  It does not matter how many green fish were caught before this fish, only that only green fish were caught after it.
Each red and blue fish has the same chance of being this last not green fish caught.  So the probability that the last not green fish is blue, given that the last fish is green is:
$$\Pr(\circ B\circ \mid \circ \circ G)= \frac{b}{r+b}$$
Thus the probability that blue fish go extinct second and green fish go extinct last is: $$\Pr(\circ BG) = \frac{b}{r+b}\frac{g}{r+b+g}$$
And by symmetry: $$\Pr(\circ GB) = \frac{g}{r+g}\frac{b}{r+b+g}$$
Finally: $$\Pr(R\circ\circ) = \Pr(\circ BG)+\Pr(\circ GB)\\ = \frac{bg(2r+b+g)}{(r+b)(r+g)(r+b+g)}$$
