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Adam1994
Three marbles are to be drawn successively from a bag that contains 4 red marbles, 2 blue marbles, and 3 green marbles. No replacement is done. What is the probability that the first marble drawn is red, the second is blue and the third is green?
You have \(4+2+3=9\) total marbles. Thus, the probability you would get a red marble is given by \(4/9\) since you can choose 4 marbles from 9. Now, we've removed a ball, so there are 8 balls left. Here, the chance to draw a blue marble would be \(2/8\) since you can choose two marbles from 8. Likewise, choosing a green marble now would be \(3/7\) for similar reasoning. Thus, your final answer should be \[{4 \over 9}\cdot{2 \over 8}\cdot{3 \over 7}\]