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A bag contains 3 red marbles, 4 white marbles and 3 black marbles. Find the probability of getting a red marble on the first draw, a black marble on the second draw, and a white marble on the third draw (a) if the marbles are drawn with replacement, (b) if the marbles are drawn without replacement.

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closed as off-topic by Song, NCh, Jyrki Lahtonen, José Carlos Santos, A. Pongrácz Feb 15 at 9:50

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    $\begingroup$ Have you made an attempt to solve the problem yourself? This looks like a homework problem. It is respectful to show your effort first. $\endgroup$ – Aditya Dua Feb 13 at 21:06
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    $\begingroup$ Not a homework question just practice for a course. I am not great at statistics or probability so I thought I'd ask. Sorry if I offended you $\endgroup$ – sidhus215 Feb 13 at 21:10
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With replacement, $\frac{3}{10}\frac{3}{10}\frac{4}{10}=\frac{9}{250}$. Without, $\frac{3}{10}\frac{3}{9}\frac{4}{8}=\frac{1}{20}$.

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With replacement: The probability of a red marble is always 3/10, the probability of a white marble is always 4/10, and the probability of a black marble is always 3/10. So the probability of RBW is simply the product of the three probabilities, i.e. ${3 \over 10} \times {3 \over 10} \times {4 \over 10} = 0.036$.

Without replacement: In this case, the probability keeps changing as you draw more marbles. On the first draw, the probability of getting a red marble is 3/10. Once you get a red marble, there are 2R, 4W, and 3B left. Now the probability of getting a black marble is 3/9 (not 3/10). Thus, the joint probability of red on first draw and black on second draw is ${3 \over 10} \times {3 \over 9}$. You can continue the same logic to see that the probability of RBW is ${3 \over 10} \times {3 \over 9} \times {4 \over 8} = 0.05$.

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