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 random variable X has the probability distribution
for the events E = {X is a prime number} and F = {X \(\leq\) 3}, then P (E\(\cup\)F)=
| \(X = x\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(P(X = x)\) | \(0.23\) | \(0.15\) | \(0.12\) | \(0.10\) | \(0.20\) | \(0.07\) | \(0.08\) | \(0.05\) |
for the events E = {X is a prime number} and F = {X \(\leq\) 3}, then P (E\(\cup\)F)=
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Use P(E union F) = P(E) + P(F) - P(E and F).
Updated On: Oct 1, 2026
- \(0.55\)
- \(0.70\)
- \(0.87\)
- \(0.78\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
\(E = \{2, 3, 5, 7\}\) (primes) and \(F = \{1, 2, 3\}\) (\(X \le 3\)). Read the probabilities from the table.
Step 2: Compute each part:
\(P(E) = 0.15 + 0.12 + 0.20 + 0.08 = 0.55\).
\(P(F) = 0.23 + 0.15 + 0.12 = 0.50\).
\(E\cap F = \{2, 3\}\), so \(P(E\cap F) = 0.15 + 0.12 = 0.27\).
Step 3: Apply the formula:
\[ P(E\cup F) = 0.55 + 0.50 - 0.27 = 0.78 \]
Step 4: Check by listing:
\(E\cup F = \{1, 2, 3, 5, 7\}\): \(0.23 + 0.15 + 0.12 + 0.20 + 0.08 = 0.78\). Option (D).
Final Answer:
P(E or F) = 0.55 + 0.50 - 0.27 = 0.78. \[ \boxed{\text{(D) }0.78} \]
\(E = \{2, 3, 5, 7\}\) (primes) and \(F = \{1, 2, 3\}\) (\(X \le 3\)). Read the probabilities from the table.
Step 2: Compute each part:
\(P(E) = 0.15 + 0.12 + 0.20 + 0.08 = 0.55\).
\(P(F) = 0.23 + 0.15 + 0.12 = 0.50\).
\(E\cap F = \{2, 3\}\), so \(P(E\cap F) = 0.15 + 0.12 = 0.27\).
Step 3: Apply the formula:
\[ P(E\cup F) = 0.55 + 0.50 - 0.27 = 0.78 \]
Step 4: Check by listing:
\(E\cup F = \{1, 2, 3, 5, 7\}\): \(0.23 + 0.15 + 0.12 + 0.20 + 0.08 = 0.78\). Option (D).
Final Answer:
P(E or F) = 0.55 + 0.50 - 0.27 = 0.78. \[ \boxed{\text{(D) }0.78} \]
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