Question:
A random variable X has the following probability distribution
X 1 2 3 4 5 6 7 8 P(X=\(x\)) 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
For the event E = { X is a prime number } and F = { X < 4 }, P(E\(\cup\)F) =
A random variable X has the following probability distribution
For the event E = { X is a prime number } and F = { X < 4 }, P(E\(\cup\)F) =
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X=\(x\)) | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
For the event E = { X is a prime number } and F = { X < 4 }, P(E\(\cup\)F) =
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Use P(E union F) = P(E) + P(F) - P(E intersect F).
Updated On: Oct 1, 2026
- \(0.87\)
- \(0.77\)
- \(0.35\)
- \(0.50\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
E = {X is prime} = {2, 3, 5, 7}. F = {X < 4} = {1, 2, 3}. The union rule is \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\).
Step 2: Probabilities
\(P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\).
\(P(F) = 0.15 + 0.23 + 0.12 = 0.50\).
\(E \cap F = \{2, 3\}\), so \(P(E \cap F) = 0.23 + 0.12 = 0.35\).
Step 3: Union
\[ P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77 \]
Option (C), 0.35, is only the probability of the common part E and F both occurring.
Final Answer:
The probability is 0.77, option (B). \[ \boxed{0.77} \]
E = {X is prime} = {2, 3, 5, 7}. F = {X < 4} = {1, 2, 3}. The union rule is \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\).
Step 2: Probabilities
\(P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62\).
\(P(F) = 0.15 + 0.23 + 0.12 = 0.50\).
\(E \cap F = \{2, 3\}\), so \(P(E \cap F) = 0.23 + 0.12 = 0.35\).
Step 3: Union
\[ P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77 \]
Option (C), 0.35, is only the probability of the common part E and F both occurring.
Final Answer:
The probability is 0.77, option (B). \[ \boxed{0.77} \]
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