A random variable X has the following probability distribution:

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:

  1. (1) $\frac{7}{12}$

  2. (2) $\frac{1}{36}$

  3. (3) $\frac{1}{6}$

  4. (4) $\frac{23}{36}$


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 $)$

$P(X>2)=K+2 K+5 K^{2}$

$=\frac{1}{6}+\frac{2}{6}+\frac{5}{36}=\frac{6+12+5}{36}=\frac{23}{36}$

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