A random variable X has the probability distr | vedclass

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(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 =

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$0.77$

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(B) The prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ are $2, 3, 5, 7$.$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.The event $F = \{X $P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.12 = 0.50$.The intersection $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.12 = 0.35$.Using the addition theorem of probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ random variable $X$ has the probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is

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$A$ random variable $X$ has the probability distribution:$X$$1$$2$$3$$4$$5$$6$$7$$8$$P(X)$$0.15$$0.23$$0.12$$0.10$$0.20$$0.08$$0.07$$0.05$For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,then $P(E \cup F)$ is