Question

A random variable $X$ has the following probability distribution:
$$\begin{array}{cccccc}X& 1& 2& 3& 4& 5\\ P\left(X\right)& K^2& 2K& K& 2K& 5K^2\end{array}$$
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 $\frac{23}{36}$

We know that $\sum _{x=1}^{5}P\left(X\right)=1$
$\Rightarrow K^2+2K+K+2K+5K^2=1$
$\Rightarrow K=-1,\frac{1}{6}\Rightarrow K=\frac{1}{6}$

$P(X>2)=P(X=3)+P(X=4)+P(X=5)$
$=K+2K+5K^2=\frac{23}{36}$