<s>[INST] probability with two variables - Taking balls out of the basket There's 3 balls in the basket - White, red and black. Three people chose one ball after the other with return.
$X$ is the various colors that got chosen.
$Y$ is the number of people that chose white.
I need to calculate $P(X+Y≤3|X-Y≥1)$.
So I got that I need to start from that getting each ball is probability of $1/3$, since it's with return.
I'm getting confused with how Can I compute Y? Can I say it's sort of uniform distribution and maybe to try by that? Meaning, that the expected value is $(a-b)/2$?
 [/INST] We can make a probability table of joint outcomes of $(X, Y)$.  Note $X \in \{1, 2, 3\}$, and $Y \in \{0, 1, 2, 3\}$.  For $X = 1$, there are only the three outcomes $$(w,w,w), (r,r,r), (b,b,b).$$  For $X = 2$, we have $\binom{3}{2}(2^3 - 2) = 18$ outcomes:
$$(r, r, w), (r, r, b), (r, w, r), (r, w, w), (r, b, r), (r, b, b), \\
  (w, r, r), (w, r, w), (w, w, r), (w, w, b), (w, b, w), (w, b, b), \\
  (b, r, r), (b, r, b), (b, w, w), (b, w, b), (b, b, r), (b, b, w).$$
For $X = 3$, we have $3! = 6$ outcomes, which are the permutations of $r, b, w$ in some order.  The total is $3^3 = 27$.
When $X = 1$, we have either $Y = 3$ with probability $1/3$ or $Y = 0$ with probability $2/3$.
When $X = 2$, we have $Y = 0$, $Y = 1$, $Y = 2$ each with probability $1/3$.
When $X = 3$, we have $Y = 1$ with probability $1$.
So we have
$$\begin{array}{c|cccc}
\Pr[X = x, Y = y] & 0 & 1 & 2 & 3 \\
\hline
1 & \frac{2}{27} & 0 & 0 & \frac{1}{27} \\
2 & \frac{2}{9} & \frac{2}{9} & \frac{2}{9} & 0 \\
3 & 0 & \frac{2}{9} & 0 & 0
\end{array}$$
The rest is simply conditioning.  Select those outcomes for which $X - Y \ge 1$, and among those, tabulate the probabilities for which $X + Y \le 3$.  then divide by the sum of the probabilities that you considered.
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