1. **Problem statement:** Chris tries to throw a ball into a basket with success probability $p=\frac{1}{3}$ each attempt, independent of others. Define $X$ as the number of attempts until success or 6 if no success in first 5 attempts. Define $Y=1$ if success in first 3 attempts, else $Y=0$. We want the conditional mass function $p_{X|Y}(x|y)$ for $x=1,2,3,4,5,6$ and $y=0,1$. 2. **Key distributions and rules:** - $X$ is a truncated geometric random variable with success probability $p=\frac{1}{3}$. - Probability of success on attempt $k$ (for $k=1$ to 5) is $P(X=k) = (1-p)^{k-1}p$. - Probability of no success in first 5 attempts is $P(X=6) = (1-p)^5$. - $Y=1$ means success in attempts 1, 2, or 3, so $P(Y=1) = 1 - (1-p)^3$. - $Y=0$ means no success in first 3 attempts, so $P(Y=0) = (1-p)^3$. 3. **Calculate unconditional probabilities $P(X=x)$:** - For $x=1$ to 5: $$P(X=x) = (1-p)^{x-1}p = \left(\frac{2}{3}\right)^{x-1} \cdot \frac{1}{3}$$ - For $x=6$: $$P(X=6) = (1-p)^5 = \left(\frac{2}{3}\right)^5$$ 4. **Calculate $P(Y=1)$ and $P(Y=0)$:** - $$P(Y=1) = 1 - (1-p)^3 = 1 - \left(\frac{2}{3}\right)^3 = 1 - \frac{8}{27} = \frac{19}{27}$$ - $$P(Y=0) = (1-p)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$$ 5. **Calculate joint probabilities $P(X=x, Y=y)$:** - For $Y=1$ (success in first 3 attempts): - $x=1,2,3$ means success at attempt $x$: $$P(X=x, Y=1) = P(X=x) = \left(\frac{2}{3}\right)^{x-1} \cdot \frac{1}{3}$$ - $x=4,5,6$ means success after 3rd attempt or no success in first 5 attempts, so $Y=0$ for these, thus: $$P(X=x, Y=1) = 0 \quad \text{for } x=4,5,6$$ - For $Y=0$ (no success in first 3 attempts): - $x=1,2,3$ means success in first 3 attempts, so $P(X=x, Y=0) = 0$ - $x=4,5$ means success at attempt 4 or 5, which is after 3 failures: $$P(X=4, Y=0) = P(X=4) = \left(\frac{2}{3}\right)^3 \cdot \frac{1}{3} = \frac{8}{27} \cdot \frac{1}{3} = \frac{8}{81}$$ $$P(X=5, Y=0) = P(X=5) = \left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{81} \cdot \frac{1}{3} = \frac{16}{243}$$ - $x=6$ means no success in first 5 attempts: $$P(X=6, Y=0) = P(X=6) = \left(\frac{2}{3}\right)^5 = \frac{32}{243}$$ 6. **Calculate conditional probabilities $p_{X|Y}(x|y) = \frac{P(X=x, Y=y)}{P(Y=y)}$:** - For $Y=1$: - $x=1$: $$p_{X|Y}(1|1) = \frac{\left(\frac{2}{3}\right)^0 \cdot \frac{1}{3}}{\frac{19}{27}} = \frac{\frac{1}{3}}{\frac{19}{27}} = \frac{1}{3} \cdot \frac{27}{19} = \frac{9}{19}$$ - $x=2$: $$p_{X|Y}(2|1) = \frac{\left(\frac{2}{3}\right)^1 \cdot \frac{1}{3}}{\frac{19}{27}} = \frac{\frac{2}{3} \cdot \frac{1}{3}}{\frac{19}{27}} = \frac{2}{9} \cdot \frac{27}{19} = \frac{6}{19}$$ - $x=3$: $$p_{X|Y}(3|1) = \frac{\left(\frac{2}{3}\right)^2 \cdot \frac{1}{3}}{\frac{19}{27}} = \frac{\frac{4}{9} \cdot \frac{1}{3}}{\frac{19}{27}} = \frac{4}{27} \cdot \frac{27}{19} = \frac{4}{19}$$ - $x=4,5,6$: $$p_{X|Y}(x|1) = 0$$ - For $Y=0$: - $x=1,2,3$: $$p_{X|Y}(x|0) = 0$$ - $x=4$: $$p_{X|Y}(4|0) = \frac{\frac{8}{81}}{\frac{8}{27}} = \frac{8}{81} \cdot \frac{27}{8} = \frac{27}{81} = \frac{1}{3}$$ - $x=5$: $$p_{X|Y}(5|0) = \frac{\frac{16}{243}}{\frac{8}{27}} = \frac{16}{243} \cdot \frac{27}{8} = \frac{432}{1944} = \frac{1}{4.5} = \frac{2}{9}$$ - $x=6$: $$p_{X|Y}(6|0) = \frac{\frac{32}{243}}{\frac{8}{27}} = \frac{32}{243} \cdot \frac{27}{8} = \frac{864}{1944} = \frac{4}{9}$$ 7. **Summary of conditional probabilities:** | x | $p_{X|Y}(x|1)$ | $p_{X|Y}(x|0)$ | |---|----------------|----------------| | 1 | $\frac{9}{19}$ | 0 | | 2 | $\frac{6}{19}$ | 0 | | 3 | $\frac{4}{19}$ | 0 | | 4 | 0 | $\frac{1}{3}$ | | 5 | 0 | $\frac{2}{9}$ | | 6 | 0 | $\frac{4}{9}$ | These 12 values fully describe the conditional mass function $p_{X|Y}(x|y)$. **Final answer:** $$ p_{X|Y}(x|y) = \begin{cases} \frac{9}{19}, \frac{6}{19}, \frac{4}{19}, 0, 0, 0 & \text{for } y=1, x=1..6 \\ 0, 0, 0, \frac{1}{3}, \frac{2}{9}, \frac{4}{9} & \text{for } y=0, x=1..6 \end{cases} $$