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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 $K = \frac{1}{6}$: $P(X > 2) = 6(\frac{1}{6})^2 + 2(\frac{1}{6}) = 6(\frac{1}{36}) + \frac{2}{6} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

$X=x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$K$	$2K$	$2K$	$3K$	$K^2$	$2K^2$	$7K^2+K$

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

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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The range of a random variable $X$ is $\{0, 1, 2\}$. If $P(X=0) = 3C^3$,$P(X=1) = 4C - 10C^2$,and $P(X=2) = 5C - 1$,then the value of $C$ is

The probability distribution of a random variable $X$ is given below:
$X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X=x)$ $0$ $K$ $2K$ $2K$ $3K$ $K^2$ $2K^2$ $7K^2+K$

Then,$P(0 < X < 5)$ is equal to:

$A$ 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:

Let a random variable $X$ take values $\{0, 1, 2, 3\}$ with $P(X=0) = P(X=1) = p$,$P(X=2) = P(X=3) = q$,and $E(X^2) = 2E(X)$. Then the value of $8p - 1$ 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