A random variable X has the following probabi | vedclass

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(B) For a probability distribution,the sum of all probabilities must be equal to $1$.$k^2 + 2k + k + 2k + 5k^2 = 1$$6k^2 + 5k = 1$$6k^2 + 5k - 1 = 0$$

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$\frac{23}{36}$

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(B) For a probability distribution,the sum of all probabilities must be equal to $1$.$k^2 + 2k + k + 2k + 5k^2 = 1$$6k^2 + 5k = 1$$6k^2 + 5k - 1 = 0$$6k^2 + 6k - k - 1 = 0$$6k(k + 1) - 1(k + 1) = 0$$(6k - 1)(k + 1) = 0$Since $P(X) \geq 0$,$k$ must be positive,so $k = \frac{1}{6}$ (as $k = -1$ is not possible).We need to find $P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)$.$P(X > 2) = k + 2k + 5k^2 = 3k + 5k^2$.Substituting $k = \frac{1}{6}$:$P(X > 2) = 3(\frac{1}{6}) + 5(\frac{1}{6})^2$$P(X > 2) = \frac{1}{2} + \frac{5}{36}$$P(X > 2) = \frac{18}{36} + \frac{5}{36} = \frac{23}{36}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

$A$ 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:

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$A$ 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: