All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{P}(R = 4) = \frac{{^{18}\!C_{4}}{^{12}\!C_{6}}}{{^{30}\!C_{10}}} \approx 0.0941$

(ii)
$\mathrm{P}(R = r) > \mathrm{P}(R = r+1)$

$\Rightarrow \frac{{^{18}\!C_r}{^{12}\!C_{10-r}}}{{^{30}\!C_{10}}} > \frac{{^{18}\!C_{r+1}}{^{12}\!C_{9-r}}}{{^{30}\!C_{10}}}$

$(\frac{18!}{r!(18-r)!})(\frac{12!}{(10-r)!(2+r)!}) > (\frac{18!}{(r+1)!(17-r)!})(\frac{12!}{(9-r)!(3+r)!})$

$(r+1)! (17-r)! (9-r)! (r+3)! > r! (18-r)! (10-r)! (r+2)!$

$\frac{(r+1)!(r+3)!}{r!(r+2)!} > \frac{(18-r)!(10-r)!}{(17-r)!(9-r)!}$

$(r+1)(r+3) > (18-r)(10-r)$

$r^2 + 4r +3 > 180 -28 +r^2$

$r > 5.53$

Since r is an integer, $r = 6$

(ii) is honestly not easy. Some students didn’t know the formula which is puzzling as the formal is given in MF15. I think it is important that students knows what formulas they are given in exams so they can utilise they properly.

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