# A random variable X has the following probability distribution:          $X: \: \: 1\; \; \: \: 2\; \;\: \: \: \: 3\; \; \: \: \: \: \: 4\: \: \: \; \; \: \: 5$ $P(X): \;K^{2}\; \; 2K\; \; K\; \; 2K\; \; 5K^{2}$ Then $P(X>2)$ is equal to:  Option: 1 Option: 2 Option: 3 Option: 4

$\\\sum P_i=1\Rightarrow 6k^2+5k=1\\\\\Rightarrow 6k^2+5k-1=0\\\\\Rightarrow k=\frac{1}{6},\:k=-1\text{ (invalid)}\\\\ Now,\,\,{P(X>2)=P(3)+P(4)+P(5)=k+2 k+5 k^{2}} \\\\ {=\frac{1}{6}+\frac{2}{6}+\frac{5}{36}=\frac{6+12+5}{36}=\frac{23}{36}}$

Correct Option 2

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