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, $\Sigma P(X)=1$
$\Rightarrow 0+K+2 K+2 K+3 K+K^{2}+2 K^{2}+7 K^{2}+K=1$
$\Rightarrow 9 K+10 K^{2}=1$
$\Rightarrow 10 K^{2}+9 K-1=0$
$\Rightarrow 10 K^{2}+10 K-K-1=0$
$\Rightarrow 10 K(K+1)-1(K+1)=0$
$\Rightarrow (K+1)(10 K-1)=0$
$\therefore K=-1, \frac{1}{10}$
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) $
$=K+2 K+2 K+3 K=8 K$
$=\frac{8}{10} [\because K=\frac{1}{10}] $