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An urn contains three balls $1$ $white$ and $2$ $black$ , a ball is drawn at random from the urn with replacement , also $X$ denotes an event such that $X=0$ means a white ball is obtained and $X=1$ means a black ball.

($X_1,X_2,X_3,X_4,X_5,X_6,X_7,X_8,X_9$) denotes a sample observed from the given event , we want the joint distribution of ($X_1,X_2,X_3,X_4,X_5,X_6,X_7,X_8,X_9$).

What I think is $X_i$ can take values as $0$ and $1$ only , thus joint distribution will have values like this - ($0,0,1,0,1,0,1,0,0$).

Now my question is wouldn't that approach come out to be clumsy ? I mean there will be a lot of cases like these - ($0,0,1,0,1,0,1,0,0$) , ($0,0,1,0,1,0,1,1,1$), ($1,1,1,0,1,0,1,0,0$)... .. etc.

Is there a better way to approach this question ? Please help !

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$$P(X_1=b_1,\dots,X_9=b_9)=\prod\limits_{n=1}^{9}\frac{b_i+1}{3}$$

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Let $S(b) = S(b_1,\dots,b_9) = \sum\limits_{i=1}^9 b_i$ , then $$ P(X_1=b_1,\dots,X_9=b_9)=(2/3)^{S(b)} (1/3)^{9-S(b)}=2^{S(b)}/3^9 $$ which is basically a simplification of barak manos's answer.

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