Question 3.EX.6: Repeat Exercise 5 but under the assumption that when a ball ......

\begin{aligned}& p_{X \mid Y}(1 \mid 3)=\frac{P\{X=1, Y=3\}}{P\{Y=3\}} \\& =\frac{P\{1 \text { white, } 3 \text { black, } 2 \text { red }\}}{P\{3 \text { black }\}} \\& =\frac{\frac{6 !}{1 ! 3 ! 2 !}\left(\frac{3}{14}\right)^1\left(\frac{5}{14}\right)^3\left(\frac{6}{14}\right)^2}{\frac{6 !}{3 ! 3 !}\left(\frac{5}{14}\right)^3\left(\frac{9}{14}\right)^3} \\& =\frac{4}{9} \\& p_{X \mid Y}(0 \mid 3)=\frac{8}{27} \\& p_{X \mid Y}(2 \mid 3)=\frac{2}{9} \\&\end{aligned}
\begin{aligned}p_{X \mid Y}(3 \mid 3) & =\frac{1}{27} \\E[X \mid Y=1] & =\frac{5}{3}\end{aligned}