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(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$

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$\left(\frac{9}{10}\right)^2$

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(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$ $10k^2 + 9k - 1 = 0$ $(10k - 1)(k + 1) = 0$ Since $P(X=x) \ge 0$,$k$ must be positive,so $k = \frac{1}{10}$. Now,we need to find $P(0 $P(0 Substituting $k = \frac{1}{10}$: $P(0 < X < 6) = 8\left(\frac{1}{10}\right) + \left(\frac{1}{10}\right)^2 = \frac{8}{10} + \frac{1}{100} = \frac{80+1}{100} = \frac{81}{100} = \left(\frac{9}{10}\right)^2$.

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

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

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

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

Find the value of $P(0 < X < 6)$.

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$A$ random variable $X$ has the following probability distribution: $x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Find the value of $P(0 < X < 6)$.