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(A) We know that the sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X) = 1$.$0 + k + 2k + 2k + 3k + k^2

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$k = \frac{1}{10}$

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(A) We know that the sum of all probabilities in a probability distribution must be equal to $1$.Therefore,$\sum P(X) = 1$.$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$Combining the terms,we get:$(k^2 + 2k^2 + 7k^2) + (k + 2k + 2k + 3k + k) = 1$$10k^2 + 9k = 1$$10k^2 + 9k - 1 = 0$Factoring the quadratic equation:$(10k - 1)(k + 1) = 0$This gives $k = \frac{1}{10}$ or $k = -1$.Since the probability $P(X)$ cannot be negative,$k$ must be positive. If $k = -1$,then $P(X=1) = k = -1$,which is impossible.Thus,the only valid value is $k = \frac{1}{10}$.

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

$X = x$	$1$	$2$	$3$	$4$
$P(X = x)$	$1/10$	$2/10$	$3/10$	$4/10$

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Determine $k$.

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$A$ random variable $X$ has the following probability distribution: $X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Determine $k$.