If a random variable X follows the Binomial distribution B(33, p)...

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

Given $3P$\left(X = 0\right) = P$\left(X = 1\right)</p><p>$3\left(C0 33\right)\left({p}^{0}\right)\left({q}^{33}\right)$= C1 33 \left({p}^{1}\right)\left({q}^{32}\right)</p><p>$3 \times 1 \times q = 33 \times p$</p><p>$q = 11p . . . \left(i\right)$</p><p>Also$p + q = 1$</p><p>Now from equation \left(i\right), p + 11p = 1$

$12p = 1$

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

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

$⇒ \dfrac{C15 33 {p}^{15}{q}^{18}}{C18 33 {p}^{18}{q}^{15}} - \dfrac{C16 33 {p}^{16}{q}^{17}}{C17 33 {p}^{17}{q}^{16}}$

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

$= 11 \times 120$

$= 1320$