Question

The probability distribution of a random variable is given below

$X=x$	0	1	2	3	4	5	6	7
$P(X=x)$	0	K	2k	2K	3K	$K^2$	$2K^2$	$7K^2+K$

Then, $P(0<x<5)$ is equal to

• A. $\frac{1}{10}$
• B. $\frac{3}{10}$
• C. $\frac{8}{10}$
• D. $\frac{7}{10}$

Answer 1

Answer: (C)

We know that, $\Sigma P(X)=1$
$\Rightarrow 0+K+2K+2K+3K+K^2+2K^2+7K^2+K=1$
$\Rightarrow 9K+10K^2=1$
$\Rightarrow 10K^2+9K-1=0$
$\Rightarrow 10K^2+10K-K-1=0$
$\Rightarrow 10K(K+1)-1(K+1)=0$
$\Rightarrow (K+1)(10K-1)=0$
$\therefore K=-1,\frac{1}{10}$
But $K>0$ as probability cannot be negative.
$\therefore K=\frac{1}{10}$
$P(0<x<5)=P(X=1)+P(X=2)$
$+P(X=3)+P(X=4)$
$=K+2K+2K+3K=8K$
$=\frac{8}{10}[\because K=\frac{1}{10}]$