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Bag of 10 blue marbles, 4 green marbles, and 6 red marbles. Find probability in one draw pulling on 1 green and 1 red marble.

The answer sheet from teacher shows the answer is 12.6%.

I need to understand how to do these sorts of problems so I can reproduce on next test.

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Interpretation 1: The marble is not replaced between draws.

You're drawing 2 marbles from 20, which can be done in $C(20,2)$ ways, and you have a success if you draw one of the 4 greens and 1 of the 6 reds, which can be done in $4\cdot 6$ ways. Then compute $C(20,2)=20\cdot 19/2=190.$ The probability is then $24/190=0.126315...$ which is to one decimal $12.6\%.$

Interpretation 2: The marble is replaced between draws. Then on the first draw you get a green with probability $4/20$, and on the second draw you get a red with probability $6/20$, giving $24/400=0.06$ (exactly $6\%$). If you draw in the opposite order, red then green, the probability is the same 6 percent. But the wording of the problem seems to mean it doesn't matter in which order the red and green are drawn, so the 6 percent must be doubled to 12 percent for the answer.

If the text is in the habit of rounding answers, it seems the first interpretation (not replacing between draws) is more likely what is meant. However one must look at how the problem is actually phrased to know which is meant.

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