I have this problem: There is a jar with 12 balls, 3 red, 4 white, and 5 blue. I take one out and replace it. Then I take one out again. I'm supposed to figure out the joint probability mass of X (# of red balls) and Y (# of white balls). I've tried drawing out the table showing all the possible combinations from X = 0, Y = 0 to X = 2, Y = 2, and filling in the probabilities, but it will not add up to 1.
For example, for X = 0, Y = 0 I have (5/12) * (5/12) = 25/144, since this is the case where I get a blue ball both times. I have calculated these odds for every possibility, yet end up with 121/144 in total. Where is the other 23/144?
$\begin{array}{r|c|c|c|}&Y=0&Y=1&Y=2\\ \hline X=0&\frac{5}{12}\times\frac{5}{12}&2\times \frac{5}{12}\times\frac{4}{12}&\frac{4}{12}\times\frac{4}{12}\\\hline X=1&2\times\frac{5}{12}\times\frac{3}{12}&2\times\frac{4}{12}\times\frac{3}{12}&0\\ \hline X=2&\frac{3}{12}\times\frac{3}{12}&0&0\end{array}$
Notice the factors of $2$. These come from the fact that when pulling out in sequence, we might have either pulled the colors in one order, or in the other order.
Adding up over the entire table, we do get $1$, as expected. The most likely explanations of your error are either from forgetting the extra factor of $2$ in some scenarios, using the wrong numerator in some scenarios, or simple arithmetic mistakes.