A random variable X has the following probabi | vedclass

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(D) The probability distribution is given as:$P(X=1)=0.15, P(X=2)=0.23, P(X=3)=0.10, P(X=4)=0.12, P(X=5)=0.20, P(X=6)=0.08, P(X=7)=0.07, P(X=8)=0.05$.

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$0.75$

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(D) The probability distribution is given as:$P(X=1)=0.15, P(X=2)=0.23, P(X=3)=0.10, P(X=4)=0.12, P(X=5)=0.20, P(X=6)=0.08, P(X=7)=0.07, P(X=8)=0.05$.Event $E = \{ X 	ext{ is a prime number} \} = \{ 2, 3, 5, 7 \}$.$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.10 + 0.20 + 0.07 = 0.60$.Event $F = \{ X $P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.10 = 0.48$.Event $E \cap F = \{ X 	ext{ is a prime number and } X $P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.10 = 0.33$.Using the addition rule for probability:$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$P(E \cup F) = 0.60 + 0.48 - 0.33 = 0.75$.

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

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

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

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

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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$$0.23$$0.10$$0.12$$0.20$$0.08$$0.07$$0.05$For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is