Question

If a random variable $X$ follows the Binomial distribution $B(33$, p) such that $3P(X=0)=P(X=1)$, then the value of $\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}$ is equal to

• A. 1320
• B. 1088
• C. $\frac{120}{1331}$
• D. $\frac{1088}{1089}$

Answer 1

Answer: (A)

$n=33$, let probability of success is $p$ and $q=1-p$
$3p(x=0)=p(x=1)$
$3.{}^{33}{C}_0(q{)}^{33}={}^{33}{C}_{1}p{q}^{32}$
$p=\frac{1}{12},q=\frac{11}{12},\frac{q}{p}=11$
$\frac{p(x=15)}{p(x=18)}-\frac{p(x=16)}{p(x=17)}$
$\frac{{}^{33}{C}_{15}{p}^{15}{q}^{18}}{{}^{33}{C}_{18}{p}^{18}{q}^{15}}-\frac{{}^{33}{C}_{16}{p}^{16}{q}^{17}}{{}^{33}{C}_{17}{p}^{17}{q}^{16}}={\left(\frac{q}{p}\right)}^3-\left(\frac{q}{p}\right)$
$=(11{)}^3-11$
$=1320$