A random variable X has the following probabi | vedclass

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(D) For a probability distribution,the sum of all probabilities must be $1$.$\Sigma P(x_i) = 0.15 + 0.23 + K + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$$0

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$0.77$

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(D) For a probability distribution,the sum of all probabilities must be $1$.$\Sigma P(x_i) = 0.15 + 0.23 + K + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$$0.88 + K = 1$$K = 1 - 0.88 = 0.12$Now,define the events:$E = \{X \mid X 	ext{ is a prime number}\} = \{2, 3, 5, 7\}$$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$$F = \{X \mid X $P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$$E \cap F = \{2, 3\}$$P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$Using the addition theorem of probability:$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$K$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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$A$ random variable $X$ has the following probability distribution:$X = x$$1$$2$$3$$4$$5$$6$$7$$8$$P(X = x)$$0.15$$0.23$$K$$0.10$$0.20$$0.08$$0.07$$0.05$For the event $E = \{X \mid X \text{ is a prime number}\}$ and the event $F = \{X \mid X < 4\}$,the probability $P(E \cup F) = $