Question:
If \( A \) and \( B \) are two events associated with an experiment such that \( P(A \cup B) = P(A \cap B) \), and \( P(A) = 1/3 \), then \( P(B) \) equals
If \( A \) and \( B \) are two events associated with an experiment such that \( P(A \cup B) = P(A \cap B) \), and \( P(A) = 1/3 \), then \( P(B) \) equals
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Equality cases in probability often force extreme values like 0 or 1.
Updated On: Apr 30, 2026
- \(0\)
- \(1/3\)
- \(2/3\)
- \(1/2\)
- \(2/5\)
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
Use identity:
\[
P(A \cup B)=P(A)+P(B)-P(A \cap B)
\]
Step 1: Substitute condition. \[ P(A)+P(B)-P(A\cap B)=P(A\cap B) \]
Step 2: Rearrange. \[ P(A)+P(B)=2P(A\cap B) \]
Step 3: Use inequality. \[ P(A\cap B) \le P(A) \] So: \[ P(A)+P(B) \le 2P(A) \] \[ \frac{1}{3}+P(B) \le \frac{2}{3} \Rightarrow P(B)\le \frac{1}{3} \] Also: \[ P(A\cap B) \le P(B) \] Combining gives only possibility: \[ P(B)=0 \]
Step 1: Substitute condition. \[ P(A)+P(B)-P(A\cap B)=P(A\cap B) \]
Step 2: Rearrange. \[ P(A)+P(B)=2P(A\cap B) \]
Step 3: Use inequality. \[ P(A\cap B) \le P(A) \] So: \[ P(A)+P(B) \le 2P(A) \] \[ \frac{1}{3}+P(B) \le \frac{2}{3} \Rightarrow P(B)\le \frac{1}{3} \] Also: \[ P(A\cap B) \le P(B) \] Combining gives only possibility: \[ P(B)=0 \]
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