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Q.
If a random variable $X$ follows the Binomial distribution $B (33$, p) such that $3 P ( X =0)= P ( X =1)$, then the value of $\frac{ P ( X =15)}{ P ( X =18)}-\frac{ P ( X =16)}{ P ( X =17)}$ is equal to
$n =33$, let probability of success is $p$ and $q =1- p$
$3 p ( x =0)= p ( x =1)$
$3. { }^{33} C _{0}( q )^{33}={ }^{33} C _{1} pq ^{32}$
$p =\frac{1}{12}, q =\frac{11}{12}, \frac{ q }{ p }=11$
$\frac{ p ( x =15)}{ p ( x =18)}-\frac{ p ( x =16)}{ p ( x =17)}$
$\frac{{ }^{33} C _{15} p ^{15} q ^{18}}{{ }^{33} C _{18} p ^{18} q ^{15}}-\frac{{ }^{33} C _{16} p ^{16} q ^{17}}{{ }^{33} C _{17} p ^{17} q ^{16}}=\left(\frac{ q }{ p }\right)^{3}-\left(\frac{ q }{ p }\right)$
$=(11)^{3}-11$
$=1320$