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(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X=x) = K + 2K + K^2 + 2K + 5K^2 = 1$.Combining l

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$\frac{1}{2}$

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(C) The sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X=x) = K + 2K + K^2 + 2K + 5K^2 = 1$.Combining like terms,we get $6K^2 + 5K - 1 = 0$.Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$.This gives $K = \frac{1}{6}$ or $K = -1$.Since probability cannot be negative,we must have $K = \frac{1}{6}$.We need to find $P(X > 2) = P(X=3) + P(X=4) + P(X=5)$.$P(X > 2) = K^2 + 2K + 5K^2 = 6K^2 + 2K$.Substituting $K = \frac{1}{6}$: $P(X > 2) = 6(\frac{1}{6})^2 + 2(\frac{1}{6}) = 6(\frac{1}{36}) + \frac{2}{6} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

Consider the probability distribution
$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$
Then the value of $P(X > 2)$ is

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Consider the probability distribution$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$Then the value of $P(X > 2)$ is