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(A) The sum of all probabilities in a probability distribution must be equal to $1$. $\sum P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1

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$\frac{19}{100}$

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(A) The sum of all probabilities in a probability distribution must be equal to $1$. $\sum P(X=x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$ $10k^2 + 9k = 1$ $10k^2 + 9k - 1 = 0$ $(10k - 1)(k + 1) = 0$ Since $k > 0$,we have $k = \frac{1}{10}$. We need to find $P(X \geqslant 6) = P(X=6) + P(X=7)$. $P(X=6) = 2k^2 = 2(\frac{1}{10})^2 = \frac{2}{100}$. $P(X=7) = 7k^2 + k = 7(\frac{1}{10})^2 + \frac{1}{10} = \frac{7}{100} + \frac{10}{100} = \frac{17}{100}$. Therefore,$P(X \geqslant 6) = \frac{2}{100} + \frac{17}{100} = \frac{19}{100}$.

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

If a random variable $X$ has the following probability distribution of $X$:

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

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

Then $P(X \geqslant 6) = $

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If a random variable $X$ has the following probability distribution of $X$: $X=x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$ Then $P(X \geqslant 6) = $