Question

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

• A. \(0.5\)
• B. \(0.77\)
• C. \(0.35\)
• D. \(0.75\)

Hint:

List the values in each event, add their probabilities and subtract the overlap.

Answer 1

The Correct Option is D

Step 1: List the values in the union directly.
$E \cup F = \{1, 2, 3, 5, 7\}$.

Step 2: Add their probabilities.
\[ 0.15 + 0.23 + 0.10 + 0.20 + 0.07 = 0.75 \]

Step 3: Cross-check using the complement.
The values not in the union are 4, 6 and 8 with probability $0.12 + 0.08 + 0.05 = 0.25$, and $1 - 0.25 = 0.75$.

Final Answer:
Option (D). \[ \boxed{0.75} \]