Question:
A random variable $X$ has the following probability distribution:
$\begin{array}{lllllll}\mathrm{X} & : & 1 & 2 & 3 & 4 & 5\end{array}$
$\mathrm{P}(\mathrm{X}) \quad: \quad \mathrm{K}^{2} \quad 2 \mathrm{~K} \quad \mathrm{~K} \quad 2 \mathrm{~K} \quad 5 \mathrm{~K}^{2}$
Then, $\mathrm{P}(\mathrm{X}>2)$ is equal to:
Correct Option: , 4
Solution:
$\sum P(K)=1 \Rightarrow 6 K^{2}+5 K=1$
$6 K^{2}+5 K-1=0$
$6 K^{2}+6 K-K-1=0$
$\Rightarrow \quad(6 K-1)(K+1)=0$
$\Rightarrow K=\frac{1}{6}(K=-1$ rejected $)$
$P(X>2)=K+2 K+5 K^{2}$
$=\frac{1}{6}+\frac{2}{6}+\frac{5}{36}=\frac{6+12+5}{36}=\frac{23}{36}$