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There is an urn with 10 red balls, 20 green balls and 30 yellow balls. I take a sample of size $n$ with replacement. $X_1$ is the number of red balls. $X_2$ is the number of green balls and $X_3$ is the number of yellow balls. Find the joint distribution of $(X_1,X_2,X_3).$

I think I that the density function of $(X_1,X_2,X_3) = P(X_1=x_i,X_2=x_{i+1},X_3=x_{i+2})$ where $i\in \{1,2,...,n\}.$

Then the distribution of $(X_1,X_2,X_3)=\sum_{i=1}^{n}P(X_1=x_i,X_2=x_{i+1},X_3=x_{i+2})$

Does that look right? Can you help me use the multinomial distribution to solve the problem I gave?

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You can calculate the joint distribution by taking the probability of a particular occurrence (counting order) and then multiplying by the number of ways that outcome can happen.

This is like in the binomial distribution, where if you have a chance of success on each trial of $p$, then the probability of three successes out of ten trials is $p^3(1-p)^7 \cdot \binom{10}{3}$

So in your case, you should figure out what is the likelihood of getting $x_1$ red, $x_2$ green, and $x_3$ yellow (counting order...akin to the figure $p^3(1-p)^7$ above) and multiply by the number of ways such a result can be ordered. (Did a red ball get drawn first? What was drawn in slot number 2? Etc.)

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  • $\begingroup$ Can you help me understand the question I had about the urn and the balls? $\endgroup$ – user66360 Nov 20 '13 at 16:40
  • $\begingroup$ Sure, I tried to add to my answer. $\endgroup$ – Eric Auld Nov 20 '13 at 16:45

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