Question

A random variable X has the probability distribution

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

for the events E = {X is a prime number} and F = {X \(\leq\) 3}, then P (E\(\cup\)F)=

• A. \(0.55\)
• B. \(0.70\)
• C. \(0.87\)
• D. \(0.78\)

Hint:

Use P(E union F) = P(E) + P(F) - P(E and F).

Answer 1

The Correct Option is D

Step 1: List the union:
Outcomes in $E\cup F$: $1, 2, 3, 5, 7$.

Step 2: Add the probabilities:
$0.23 + 0.15 + 0.12 + 0.20 + 0.08$.

Step 3: Total:
$= 0.78$, option (D). Option (A) $0.55$ is only $P(E)$, which leaves out the outcome $1$.

Final Answer:
The probability is 0.78. \[ \boxed{\text{(D) }0.78} \]