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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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3 Answers 3

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

$$\frac{2\cdot3\cdot4}{\binom93}=\frac{24}{84}=\frac27\;.$$

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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