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Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls.

Suppose also that balls are drawn from the box one at a time at random.

What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.

The answer I was given is $\dfrac{7}{12}$ and a general equation is: $$ \dfrac{b g}{1-b}+\dfrac{b g}{1-g} $$ where $$ g=\dfrac{20}{60},b=\dfrac{30}{60} $$ but why?

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2  
Your claimed answer makes no sense. Even if we draw only one ball out of the $60$ total, the probability that this ball is red will be only $10/60 = 1/6$, which is much smaller than $7/12$; and the probability of getting more than one red ball for multiple draws without replacement cannot possibly exceed $1/6$. –  heropup Jun 19 at 6:02
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Perhaps he means the probability that all 10 red balls are selected before all 20 greens or all 30 blues? Not that the probability of that happening is necessarily 7/12. –  Avraham Jun 19 at 6:05

3 Answers 3

up vote 10 down vote accepted

Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.

The probability that the last ball is blue and that the last green comes after the last red is $\dfrac{30}{10+20+30}\times \dfrac{20}{10+20} =\dfrac{1}{3}$ or more generally $b \times \dfrac{g}{1-b}$.

The probability that the last ball is green and that the last blue comes after the last red is $\dfrac{20}{10+20+30}\times \dfrac{30}{10+30} =\dfrac{1}{4}$ or more generally $g \times \dfrac{b}{1-g}$.

So the probability that all reds are drawn before the final blue and final green are drawn is $\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}$ or more generally $ \dfrac{bg}{1-b} + \dfrac{bg}{1-g}$.

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Edit

The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding.

There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $\binom{60}{10}.$ Thus the probability is

$$\frac{1}{\binom{60}{10}}\approx 1.3263 \cdot 10^{-11}.$$

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this is not the right answer. –  LCFactorization Jun 19 at 7:58
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It is the answer to a different question, answering "what is the probability that the last red ball is drawn before any blue balls or green balls are drawn". The original question is closer to "what is the probability that the last red ball is drawn before the last blue ball and last green ball are drawn". –  Henry Jun 19 at 8:41
    
You are right, I misunderstood the question. –  mfl Jun 19 at 12:25
    
I edited the question to make it clearer. –  Avraham Jun 19 at 16:30

You might be able to prove it by induction. Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens. It is easy to check your formula for $P(1,1,1)$. Note also that $P(R,B,0)=0$.

Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+G\le N$. Then $$P(R+1,B,G)=\frac{R+1}{R+1+B+G}P(R,B,G)+\frac{B}{R+1+B+G}P(R+1,B-1,G)+\frac{G}{R+1+B+G}P(R+1,B,G-1)$$ and now you can check the formula is true when $R+B+G=N+1$.

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Your "note that $P(R, B, 0) = 0$" should start all alarms ringing: as OP isn't interested in blue/green balls, their colors are irrelevant, and $P(R, B, G) = P(R, B + G, 0)$ –  vonbrand Jun 19 at 8:01
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No, I think that if you have run out of green balls, then you didn't run out of red first. In the same way, I think $P(0,B,G)=1$ –  Michael Jun 19 at 9:32

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