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can someone solve this example?

An urn contains 2 Red marbles, 3 White marbles and 4 Blue marbles. You reach in and draw out 3 marbles at random (without replacement). What is the probability that you will get one marble of each color? What is the answer if you draw out the marbles one at a time with replacement?

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There are $9$ marbles in the urn, so there are $\binom93$ different sets of $3$ marbles that you could draw without replacement. How many contain one marble of each color? To build such a set, you could pick either of the $2$ red marbles, any one of the $3$ white marbles, and any one of the $4$ blue marbles, so there are $2\cdot3\cdot4$ such sets. Each of the $\binom93$ sets is equally likely to be drawn, and $2\cdot3\cdot4$ of them are ‘successes’, so the probability of success is


If you draw with replacement, however, you can potentially draw the same marble twice, and you also have to take into account the order of the draws. On each of your $3$ draws you can get any of the $9$ marbles, so there are $9^3$ possible sequences of $3$ marbles that you can draw, and they’re all equally likely. How many of them contain one marble of each color? As in the first problem, there are $2\cdot3\cdot4=24$ different sets of $3$ marbles that will work, but each of them can be drawn in several different orders to give several different successful sequences of draws. In fact $3$ different objects can be arranged in $3!=6$ different orders, so each of those $24$ sets of $3$ marbles of $3$ different colors can be drawn in $6$ different orders. Thus, there are actually $6\cdot24$ successful sequences of $3$ draws. The final probability of success when you draw with replacement is therefore ... ?

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We do the "with replacement" problem. The event "we got a red, a white, and a blue" can happen in various orders.

We find the probability of red then white then blue. This is $\dfrac{2}{9}\cdot \dfrac{3}{9}\cdot \dfrac{4}{9}$. For since we are drawing with replacement, the results on the three picks are independent.

We can end up having obtained a red, a white, and a blue in various other ways, such as blue then red then white. This has the same probability as red then white then blue. In total there are $3!$ ways we can end up having drawn one of each colour. So the required probability is $$3!\left(\frac{2}{9}\cdot\frac{3}{9}\cdot \frac{4}{9} \right).$$

The same probabilistic approach works for sampling without replacement. The probability of red then white then blue is $\dfrac{2}{9}\cdot \dfrac{3}{8}\cdot \dfrac{4}{7}$.

For blue then red then white, we get $\dfrac{4}{9}\cdot \dfrac{2}{8}\cdot \dfrac{3}{7}$. The denominators do not change, and the numerators are permuted. So the required probability is $$3!\left(\frac{2}{9}\cdot\frac{3}{8}\cdot \frac{4}{7} \right).$$

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$$n(s)=\binom93$$ $$n(A)= \binom 31 \binom 21 \binom41$$ $$P(A)=n(A)/n(s)$$

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