# Need help with an algebra probability problem and need steps shown please

Problem:

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.

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

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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.