# Probability of selecting from an Urn without counting

I have the following trivial problem:

An urn has $6$ balls, $1$ red, $1$ blue and $4$ green. If $3$ balls are selected at random, what is the probability of them being the combination of $1$ red and $2$ green.

The problem can be trivially solved, by the following counting method:

$$\binom{4}{2}/\binom{6}{3}=3/10$$

Where number of combinations that contain only 1 red ball over the total number of combinations of size 3.

My question is that I'm trying to solve the problem using the more basic probability technique (not sure of the correct wording) I've constructed the following 2 statements:

$$Pr(1R \ 2G)=1/6\cdot3/5\cdot3/4=1/10$$ or $$Pr(1R \ 2G)=1/6\cdot(1-1/5)\cdot(1-1/4)=1/10$$

Neither of them determine the correct result. I believe I'm missing a crucial point, I would prefer hints or suggestions rather than a complete answer if possible.

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I believe that you meant to write $$\binom42/\binom63\;.$$ –  Brian M. Scott Nov 19 '12 at 5:47
@BrianM.Scott yes indeed! fixed, thank-you. –  Switzy Nov 19 '12 at 5:49
You’re welcome. –  Brian M. Scott Nov 19 '12 at 5:50

HINT: Note that $$\frac16\cdot\frac45\cdot\frac34=\frac1{10}$$ is the probability that the first ball drawn is red and the last two are green. What if the red ball is second or third?
Ah yes, there's 3 ways it can happen $$p1=1/6\cdot3/5\cdot3/4=1/10$$ or $$p2=4/6\cdot1/5\cdot3/4=1/10$$ or $$p3=4/6\cdot3/5\cdot1/4=1/10$$ $$p1+p2+p3=3/10$$ The penny has dropped thank-you! –  Switzy Nov 19 '12 at 5:59