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The value of $P (x_{i} > 2)$ for the given probability distribution is
X 1 2 3 4 5
P(X) $k^{2}$ 2k k 2k $5 k^{2}$

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$\frac{1}{36}$

$\frac{23}{36}$

$\frac{1}{6}$

$\frac{7}{12}$

answer is B.

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Detailed Solution

$\sum_{9 = 1}^{5} P (X) = 1 \Rightarrow (6 k - 1) (k + 1) = 0 \Rightarrow k = \frac{1}{6}$

$∴ P (X > 2) = k + 2 k + 5 k^{2} = \frac{23}{36}$

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