Question

A random variable $X$has the following probability distribution:

$X$	$1$	$2$	$3$	$4$	$5$
$P\left(X\right)$	$K^2$	$2K$	$K$	$2K$	$5K^2$

Then $P(X>2)$ is equal to:

• A. $\frac{7}{12}$
• B. $\frac{23}{36}$
• C. $\frac{1}{36}$
• D. $\frac{1}{6}$

Answer 1

The correct option is B

$\frac{23}{36}$

Explanation for the correct option:

Find the value of $P(X>2)$:

As we know,

$\sum _{x=1}^{5}P\left(X\right)=1$

$\Rightarrow$$K^2+2K+K+2K+5K^2=1$

$\Rightarrow$ $6K^2+5K-1=0$

$\Rightarrow$ $6K(K+1)-(K+1)=0$

$\Rightarrow$ $(K+1)(6K-1)=0$

$\Rightarrow$$K=-1orK=1/6$

$K$ cannot be negative.

$\begin{array}{rcl}\therefore P(X& >& 2)=P(X=3)+P(X=4)+P(X=5)\end{array}$

$\begin{array}{rcl}& =& K+2K+5K2\\ & =& \frac{1}{6}+2\left(\frac{1}{6}\right)+5{\left(\frac{1}{6}\right)}^2\\ & =& \frac{\mathbf{23}}{\mathbf{36}}\end{array}$

Hence, Option ‘B’ is Correct.