probability with two variables - Taking balls out of the basket

Question

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 $(a-b)/2$?

Answer 1

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.

Answer 2

There's a total of $3^3$ ways of selecting 3 balls (all balls and all ppl are different).

Given that $X$ is at most $3$ and $Y$ is at most $3$, $X-Y \geq1$ can mean two cases: $$ 1) X-Y=1 $$ if either 2 colors/1 white (WR, WB) or 1 color/0 white (R,B) were chosen. $$ 2) X-Y=2 $$ if either 3 colors/one white (WRB) or 2 colors/0 white (RB). The difference can't be 3, because you can't have all 3 colors selected without selecting white.

For $X+Y$, you have 3 cases: $$ 1)X+Y=1 $$ if (R,B) were chosen $$ 2)X+Y=2 $$ if either one white (W) or (R,B) were chosen and $$ 3)X+Y=3 $$ if either (W,B) or (W,R) were chosen.

Can you put together these cases assuming all sampling was independent and withe replacement?