Question

The probability distribution of random variable $X$ is given by:
$\begin{array}{cccccc}X& 1& 2& 3& 4& 5\\ P\left(X\right)& K& 2K& 2K& 3K& K\end{array}$
Let $p=P(1<X<4|X<3)$. If $5p=\lambda K$, then $\lambda$ is equal to

Answer 1

$p=P\left(\frac{1<X<4}{X<3}\right)=\frac{P\left(\right(1<X<4)\cap (X<3\left)\right)}{P(X<3)}$
$=\frac{P(X=2)}{P(X<3)}=\frac{2K}{K+2K}=\frac{2}{3}$
Also, $K+2K+2K+3K+K=1$
$\Rightarrow K=\frac{1}{9}$
Now, $5p=\lambda K$
$\Rightarrow 5\times \frac{2}{3}=\frac{1}{9}\lambda \Rightarrow \lambda =30$