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(B) The event $E$ is defined as $X$ being a prime number. The prime numbers in the given set are $\{2, 3, 5, 7\}$.$P(E) = P(2) + P(3) + P(5) + P(7) =

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$0.77$

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

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

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

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

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