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(A) For any probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_i) = 10k + 9k + 8k + 8k + 6k + 5k + 4k + 3k + k = 54k = 1$

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$\frac{2}{3}$

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(A) For any probability distribution,the sum of all probabilities must be $1$.$\sum P(X = x_i) = 10k + 9k + 8k + 8k + 6k + 5k + 4k + 3k + k = 54k = 1$.Therefore,$k = \frac{1}{54}$.The event $A$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,which are $\{2, 3, 5, 7\}$.$P(A) = P(2) + P(3) + P(5) + P(7) = 9k + 8k + 6k + 4k = 27k$.The event $B$ consists of values $x_i > 5$,which are $\{6, 7, 8, 9\}$.$P(B) = P(6) + P(7) + P(8) + P(9) = 5k + 4k + 3k + k = 13k$.The intersection $A \cap B$ consists of values that are both prime and greater than $5$,which is $\{7\}$.$P(A \cap B) = P(7) = 4k$.Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = 27k + 13k - 4k = 36k$.Substituting $k = \frac{1}{54}$,we get $P(A \cup B) = 36 	imes \frac{1}{54} = \frac{36}{54} = \frac{2}{3}$.

$X = x_i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$P(X = x_i)$	$10k$	$9k$	$8k$	$8k$	$6k$	$5k$	$4k$	$3k$	$k$

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

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$A$ random variable $X$ has the following probability distribution:$X = x_i$$1$$2$$3$$4$$5$$6$$7$$8$$9$$P(X = x_i)$$10k$$9k$$8k$$8k$$6k$$5k$$4k$$3k$$k$where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events,then $P(A \cup B) = $