# What is the probability of getting a certain combination of marbles in $3$ draws? [closed]

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.

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

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• Have you made an attempt to solve the problem yourself? This looks like a homework problem. It is respectful to show your effort first. – Aditya Dua Feb 13 at 21:06
• 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 – sidhus215 Feb 13 at 21:10

With replacement, $$\frac{3}{10}\frac{3}{10}\frac{4}{10}=\frac{9}{250}$$. Without, $$\frac{3}{10}\frac{3}{9}\frac{4}{8}=\frac{1}{20}$$.
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 $${3 \over 10} \times {3 \over 9} \times {4 \over 8} = 0.05$$.