Question

The probability distribution of random variable X is given by:

X	1	2	3	4	5
P(X)	K	2K	2K	3K	K

Let $p=P(1<X<4\mid X<3)$. If $5p=\lambda K$, then $\lambda$ equal to_____

Answer 1

Answer: 30

$\sum P(X)=1\Rightarrow k+2k+2k+3k+k=1$
$\Rightarrow k=\frac{1}{9}$
Now, $p=P\left(\frac{kX<4}{X<3}\right)=\frac{P(X=2)}{P(X<3)}$
$=\frac{\frac{2k}{9k}}{\frac{k}{9k}+\frac{2k}{9k}}=\frac{2}{3}$
$\Rightarrow p=\frac{2}{3}$
Now, $5p=\lambda k$
$\Rightarrow (5)\left(\frac{2}{3}\right)=\lambda (1/9)$
$\Rightarrow \lambda =30$