\begin{align} 1 &= \sum_{x=0}^1 \sum_{y=0}^3 \left\{ k(x+y) + \frac{1}{16} \right\} \\ &= k \left\{ (0+1) \cdot 4 + 2 \cdot (0+1+2+3) \right\} + 8 \cdot \frac{1}{16} \\ &= 16k + \frac{1}{2} \\ \therefore \ \ k &= \frac{1}{32} \end{align}
\begin{align} P(X=x) &= \sum_{y=0}^3 \left\{ \frac{1}{32} (x+y) + \frac{1}{16} \right\} \\ &= \frac{1}{8} x + \frac{7}{16} \\ P(Y=y) &= \sum_{x=0}^1 \left\{ \frac{1}{32} (x+y) + \frac{1}{16} \right\} \\ &= \frac{1}{16} y + \frac{5}{32} \end{align}
例えば、 $P(X=0, Y=0) \ne P(X=0) P(Y=0)$ なので、 $X$ と $Y$ は独立でない。
\begin{align} P(X=1 | Y=2) &= \frac{P(X=1,Y=2)}{P(Y=2)} \\ &= \frac{5}{9} \\ P(X=1, Y \geq 2) &= P(X=1, Y=2) + P(X=1, Y=2) \\ &= \frac{9}{32} \end{align}