If a random variable X follows the Binomial distribution B(33, p) such that
$3P(X = 0) = P(X = 1)$, then the value of ${{P(X = 15)} \over {P(X = 18)}} - {{P(X = 16)} \over {P(X = 17)}}$ is equal to :
Solution
$3 P(X=0)=P(X=1)$
<br/><br/>
$$
\begin{aligned}
&3 \cdot{ }^{n} C_{0} P^{0}(1-P)^{n}={ }^{n} C_{1} P^{1}(1-P)^{n-1} \\\\
&\frac{3}{n}=\frac{P}{1-P} \Rightarrow \frac{1}{11}=\frac{P}{1-P} \\\\
&\Rightarrow 1-P=11 P \\\\
&\Rightarrow P=\frac{1}{12}
\end{aligned}
$$
<br/><br/>
$\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \frac{{ }^{33} C_{15} P^{15}(1-P)^{18}}{{ }^{33} C_{18} P^{18}(1-P)^{15}}-\frac{{ }^{33} C_{16} P^{16}(1-P)^{17}}{{ }^{33} C_{17} P^{17}(1-P)^{16}} \\\\
&\Rightarrow\left(\frac{1-P}{P}\right)^{3}-\left(\frac{1-P}{P}\right) \\\\
&\Rightarrow \quad 11^{3}-11=1320
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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