Joint probability? Suppose that $3$ balls are chosen with replacement from an urn consisting of $5$ white and $8$ red balls. Suppose that the white balls are numbered, and let $Y_i$ equal $1$ if the $i$-th white ball is selected and $0$ otherwise. Find the joint probability mass function of $Y_1,Y_2$.
My attempt is: let $Y_1$ be  the event that the white ball number $1$ is chosen and $Y_2$ be the event that the white ball number $2$ is chosen
$$p(0,0)=11*11*11/13*13*13=1331/2197$$
$$p(1,0)=1*11*11/13*13*13=121/2197$$
$$p(0,1)=1*11*11/13*13*13=121/2197$$
$$p(1,1)=1*1*11/13*13*13=11/2197$$
but the sum is not $1$.
could someone help me to understand my mistake?
 A: Let $W_1$ denote the first white ball, $W_2$ the second.  We remark that there are $11$ balls aside from those two, thus the probability of choosing a neutral ball is $\frac {11}{13}$.  We will let $X$ denote a generic "neutral" choice.
$P(0,0)$:  We need each choice to be neutral so $$P(0,0)=\left( \frac {11}{13}\right)^3=\frac {1331}{2197}$$
$P(1,0)$:  here we have to distinguish a couple of cases.
Case I:  you choose $W_1$ exactly once.  Then there are $3$ places to locate the special ball, so $\frac {3\times 11^2}{13^3}$.
Case II:  you choose $W_1$ exactly twice.  Then there are $3$ places to locate the neutral ball so $\frac {3\times 11}{13^3}$.
Case III:  You choose $W_1$ three times. $\frac 1{13^3}$
Thus $$P(1,0)=\frac {3\times 11^2+3\times 11+1}{13^3}=\frac {397}{2197}$$
$P(0,1)=P(1,0)$ by symmetry.
$P(1,1)$:  you have to distinguish two cases.
Case I:  the third ball is neutral.  Then there are $3$ ways to place $W_1$ and then $2$ ways to place $W_2$ and then $11$ choices for $X$ so $\frac  {6\times 11}{13^3}$
Case II:  The third ball is one of $W_1,W_2$.  Then there are $6$ cases (two choices for the duplicate and three ways to place the odd man out).  Thus $\frac 6{13^3}$.
Thus $$P(1,1)=\frac {6\times 11+6}{13^3}=\frac {72}{2197}$$
Consistency check:  these should add to $1$.  Indeed $$1331+397+397+72=2197$$
