Question

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

Answer: (B)

$\sum P\left(X\right) = 1 \Rightarrow K^{2} + 2K + K + 2K + 5K^{2} = 1$
$\Rightarrow 6K^{2} + 5K - 1 = 0 \Rightarrow \left(6K - 1\right) \left(K + 1\right) = 0$
$\Rightarrow K = -1 \left(rejected\right) \Rightarrow K = \frac{1}{6}$
$P\left(X > 2\right) = K + 2K + 5K^{2} =\frac{23}{36}$