The value of $K$ , if the probability distribution of a discrete random variable $x$ is
$X = x_{1} : 123$
$p (X = x_{1}) : \frac{1}{k^{2}} \frac{2}{k^{2}} \frac{3}{k^{2}}$

\sqrt{6}

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- \sqrt{6}

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\pm \sqrt{6}

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Solution

The correct option is C $\pm \sqrt{6}$
$3 \sum 1 P (X = x) = 1$
$\frac{1}{K^{2}} + \frac{2}{K^{2}} + \frac{3}{K^{2}} = 1$
hence $K^{2} = 6$
$K = \pm \sqrt{6}$

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