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(D) Step $1$: Identify the outcomes for events $E$ and $F$.$E = \{ X 	ext{ is a prime number} \} = \{2, 3, 5\}$.$F = \{ X < 4 \} = \{1, 2, 3\}$.Step

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(D) Step $1$: Identify the outcomes for events $E$ and $F$.$E = \{ X 	ext{ is a prime number} \} = \{2, 3, 5\}$.$F = \{ X Step $2$: Identify the union $E \cup F$.$E \cup F = \{1, 2, 3, 5\}$.Step $3$: Calculate the probability $P(E \cup F)$.$P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5)$.$P(E \cup F) = 0.15 + 0.20 + 0.25 + 0.10 = 0.70$.Note: The provided options do not contain $0.70$. Re-evaluating the question logic, if $E \cup F$ is calculated as $P(E) + P(F) - P(E \cap F)$:$P(E) = 0.20 + 0.25 + 0.10 = 0.55$.$P(F) = 0.15 + 0.20 + 0.25 = 0.60$.$E \cap F = \{2, 3\}$, so $P(E \cap F) = 0.20 + 0.25 = 0.45$.$P(E \cup F) = 0.55 + 0.60 - 0.45 = 0.70$.

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$A$ random variable $X$ has the following probability distribution:
$X = 1, P(X) = 0.15$
$X = 2, P(X) = 0.20$
$X = 3, P(X) = 0.25$
$X = 4, P(X) = 0.30$
$X = 5, P(X) = 0.10$
For the event $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cup F)$.

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$A$ random variable $X$ has the following probability distribution:$X = 1, P(X) = 0.15$$X = 2, P(X) = 0.20$$X = 3, P(X) = 0.25$$X = 4, P(X) = 0.30$$X = 5, P(X) = 0.10$For the event $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cup F)$.