A bag contains 5 red marbles and 7 green marbles. Two marbles are drawn randomly one at a time, and without replacement. Find the probability of picking a red and a green, without order.

This is how I attempted the question: I first go $P(\text{Red})= 5/12$ and $P(\text{Green})= 7/11$ and multiplied the two: $$\frac{7}{11}\times \frac{5}{12}= \frac{35}{132}$$ Then I got $P(\text{Green})= 7/12$ and $P(\text{Red})= 5/11$ $\implies$ $$\frac{5}{11} × \frac{7}{12}= \frac{35}{132}$$ So I decided that $$P(\text{G and R}) \;\text{ or }\; P(\text{R and G}) =\frac{35}{132} + \frac{35}{132} =\frac{35}{66}$$ Is this correct?

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This is how I attempted the question: I first got P(Red)= 5/12 and P(Green)= 7/(11 ) and multiplied the two: 7/11× 5/12= 35/132 THen I got P(Green)= 7/12 and P(Red)= 5/11=> 5/11 × 7/12= 35/132 So I decided that P(G and R) or P(R and G) =35/132 + 35/132 =35/66 Is this correct? – Sylvester Aug 10 '13 at 15:07
I'd just like to thank you for explaining how you attempted the problem when asking for help. Too many people are lazy and just want someone to give them the answer. By explaining what you've tried, people can tell you whether your thought process is on the right track (and they tend to be more willing to help). Keep it up! – Zev Chonoles Aug 10 '13 at 15:09
You took care of red and then green. Do the same for green and then red, and add. Added: I wrote this while you were still editing. Yes, you got it right. – André Nicolas Aug 10 '13 at 15:10
@Sylvester: Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. – Zev Chonoles Aug 10 '13 at 15:12

Very nice and successful attempt. You recognized that there are two ways once can draw a red and green marble, given two draws: Red then Green, or Green then Red. You took into account that the marbles are not replaced. And your computations are correct: you multiplied when you needed to multiply and added when you needed to add:

$$\left[P(\text{1. Red}) \times P(\text{2. Green})\right]+ \left[P(\text{1. Green}) \times P(\text{2. Red})\right]$$

Your method and result are correct.

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No, not at all, @Babak! – amWhy Aug 10 '13 at 16:57
@amWhy: Needs another TU! +1 – Amzoti Aug 11 '13 at 0:05

Your method is correct. For more complicated problems of the same general kind, one might take a slightly different approach.

Imagine that the $12$ marbles are distinct, they have different driver license numbers. There are $\binom{12}{2}$ equally likely ways to choose $2$ marbles from the $12$.

There are $\binom{5}{1}$ ways to choose a red marble, and $\binom{7}{1}$ ways to choose a green marble. Thus there are $\binom{5}{1}\binom{7}{1}$ ways to choose a red and a green. It follows that our probability is $$\frac{\binom{5}{1}\binom{7}{1}}{\binom{12}{2}}.$$ Harder! However, suppose we have $25$ marbles, $10$ red and $15$ green. We choose (without replacement) $8$ marbles. What is the probability that we get $3$ red and $5$ green? The same analysis shows that the probability is $$\frac{\binom{10}{3}\binom{15}{5}}{\binom{25}{8}}.$$

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