We know that $p=$success, $q=$failure

Given $3P\left(X=0\right)=P\left(X=1\right)$

$3\left({}^{33}C_0\right)\left(p^0\right)\left(q^{33}\right)={}^{33}C_1\left(p^1\right)\left(q^{32}\right)$

$3\times 1\times q=33\times p$

$q=11p ... \left(\mathrm{i}\right)$

Also $p+q=1$

Now from equation $\left(\mathrm{i}\right), p+11p=1$

$12p=1$

$p=\frac{1}{12}, q=\frac{11}{12}$

Now $\frac{P\left(X=15\right)}{P\left(X=18\right)}-\frac{P\left(X=16\right)}{P\left(X=17\right)}$

$\Rightarrow \frac{{}^{33}C_{15}{p}^{15}{q}^{18}}{{}^{33}C_{18}{p}^{18}{q}^{15}}-\frac{{}^{33}C_{16}{p}^{16}{q}^{17}}{{}^{33}C_{17}{p}^{17}{q}^{16}}$

$=\frac{q^3}{p^3}-\frac{q}{p}=\frac{q}{p}\left({\left(\frac{q}{p}\right)}^2-1\right)=\frac{11}{1}\left({11}^2-1\right)$

$=11\times 120$

$=1320$