(i) $\mathrm{P}(R=1)$
$$
\begin{array}{l}
=\frac{{ }^{17} \mathrm{C}_1{ }^{11} \mathrm{C}_{11}}{{ }^{28} \mathrm{C}_{12}} \\
=\frac{1}{1789515} [B1]\\
\mathrm{P}(R=2) \\
=\frac{{ }^{17} \mathrm{C}_2{ }^{11} \mathrm{C}_{10}}{{ }^{28} \mathrm{C}_{12}} \\
=\frac{88}{1789515}>\mathrm{P}(R=1)(\text { shown })
\end{array}
$$
[A1]
(ii) Given $\mathrm{P}$ (4 rectangles chosen)
$=15 \mathrm{P}(3$ rectangles chosen $)$
$$
\begin{aligned}
{\left[\frac{{ }^{17+r} \mathrm{C}_4 \times{ }^{11} \mathrm{C}_8}{{ }^{28+r} \mathrm{C}_{12}}\right] } &=15\left[\frac{{ }^{17+r} \mathrm{C}_3 \times{ }^{11} \mathrm{C}_9}{{ }^{28+r} \mathrm{C}_{12}}\right][\mathrm{B} 1] \\
{ }^{17+r} \mathrm{C}_4 \times{ }^{11} \mathrm{C}_8 &=15\left[{ }^{17+r} \mathrm{C}_3 \times{ }^{11} \mathrm{C}_9\right] \\
\frac{(17+r) !}{4 !(17+r-4) !} \times 165 &=15\left[\frac{(17+r) !}{3 !(17+r-3) !} \times 55\right] \\
\frac{165}{24(13+r) !} &=\frac{825}{6(14+r) !} \\
\frac{(14+r) !}{(13+r) !} &=\frac{825}{6} \times \frac{24}{165} [B1]\\
\frac{(14+r) \times(13+r) !}{(13+r) !} &=20 \\
14+r &=20 \\
r &=6
\end{aligned}
$$
[A2]
