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The question tell us that we draw two balls at random (without replacement). If at least one of the two balls is red, we draw one more ball and stop. Otherwise, we draw two more balls without replacement.

I see there are a number of scenarios here where the last ball could be Red:

(1) We choose two balls, 1 is Red, the other is notRed. We pick a third ball, it is Red. [RR'R which is identical to R'RR]

(2) We choose two balls, both are Red. We pick a third ball, it is also Red. [RRR]

(3) We choose two balls, neither are Red. We choose another two, one is Red, the other is notRed. [R'R'R'R]

(4) We choose two balls, neither are Red. We choose another two, both are Red. [R'R'RR]

The difficulty here is in scenario (3) in particular as we must have 3 instances of notRed [R'] whilst there are only 2 Green balls (limits options).

Not sure where exactly to start with solving this problem.

The exhaustive list of options (I think) is:

RRR, RBR, RGR, BRR, GRR, BBBR, BBGR, BBRR, GGBR, GGRR, BGBR, BGGR, BGRR, GBRR, GBBR, GBGR

However surely there must be a more logical solution.

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    What is meant by "the last ball"? Do you mean the last ball drawn or something else? In any case...three things can happen at the first step, according to how many red balls are drawn. Compute the probability for each. Then compute the probability of the desired event in each case. Green vs Black has no meaning...we only care about Red vs non-Red. – 2017-01-17
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    Yes, the last ball means the last ball drawn - I understand that there is no distinction between Green/Black necessarily however the difficulty lies in the fact that there are only 2 green balls as explained above. – 2017-01-17
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    Not following. Ignore Green, Black. We have $4$ Red balls and $5$ non-Red balls. Your method gets complicated because you carefully keep track of Green and Black but that isn't necessary. – 2017-01-17
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    For example: the probability that step I yields $0$ Red balls is $\frac 59\times \frac 48$. The probability that it yields $2$ Red balls is $\frac 49\times \frac 38$. You can get the probability that it yields exactly $1$ Red either directly or by subtraction. – 2017-01-17
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    Thank you. Understood. As mentioned above, outcomes R'RR and RR'R are identical in terms of probability (as are R'R'R'R and R'R'RR') - do I count these twice ? – 2017-01-17
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    I wouldn't look to outcomes that way (too easy to make careless errors). I'd use the law of total probability. Your answer is $\sum p(n)q(n)$. Where $p(n)$ is the probability that step I yields exactly $n$ Reds and $q(n)$ is the probability that the final draw is Red given that you drew exactly $n$ Reds in Step I. I already computed $p(n)$ in a prior comment, $q(n)$ is straightforward. – 2017-01-17
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    To your question in your last comment: both of those scenarios contribute to the $n=1$ case in my formula. Yes, you need to consider both of them (if you do it by enumerating paths). – 2017-01-17
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    Perfect, this is really helpful. To make sure I have understood you correctly, my solution should look something like this: $(\frac{5}{9}*\frac{4}{8}*\frac{4}{7}) + (\frac{4}{9}*\frac{3}{8}*\frac{2}{7}) + (\frac{5}{9}*\frac{3}{7}) = \frac{4}{9}$ – 2017-01-17
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    That looks perfect! – 2017-01-17
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    Thank you. As a follow up: The question then asks for the probability that the last ball is Red given that the first ball was black i.e $P(R_n | B_1)$ We have just found that $P(R_n) = \frac{4}{9}$ and $P(B_1)$ is intuitive... However it doesn't make sense to use Bayes Theorem and I can't think of any other solution. – 2017-01-17
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    The rules are otherwise the same? So there is probability $\frac 48$ that the second draw was Red, prob $\frac 48$ that it was non-Red. Now proceed as before. – 2017-01-17
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    Yes - same rules. In looking for a shortcut way of solving this I overlooked the fact that the prior methodology can be applied here also. Thank you! – 2017-01-17
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    Yeah, I don't see a shortcut. – 2017-01-17

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