A random variable X has the following probabi | vedclass

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\Sigma P(X) = 1$.$(k-1) + 3k + k + 3k + 3k^2 + k^2 + (k^2+k

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$ \frac{1}{5} $

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\Sigma P(X) = 1$.$(k-1) + 3k + k + 3k + 3k^2 + k^2 + (k^2+k) = 1$Combining like terms:$5k^2 + 9k - 1 = 1$$5k^2 + 9k - 2 = 0$Factoring the quadratic equation:$5k^2 + 10k - k - 2 = 0$$5k(k+2) - 1(k+2) = 0$$(5k-1)(k+2) = 0$This gives $k = \frac{1}{5}$ or $k = -2$.Since the probability $P(X)$ cannot be negative,$k$ must be positive. For $k = -2$,$P(X=1) = -2-1 = -3$,which is impossible.Therefore,$k = \frac{1}{5}$.

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

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

$A$ random variable $X$ has the following probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $k-1$ $3k$ $k$ $3k$ $3k^2$ $k^2$ $k^2+k$

Then the value of $k$ is:

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$A$ random variable $X$ has the following probability distribution:$X$$1$$2$$3$$4$$5$$6$$7$$P(X)$$k-1$$3k$$k$$3k$$3k^2$$k^2$$k^2+k$Then the value of $k$ is: