Question

Let A and B be two events with probabilities \( P(A) = \frac{3}{4} \) and \( P(B) = \frac{1}{3} \); then which of the following options is true?

• A. \( P(A \cap B) \geq \max\left( \frac{3}{4}, \frac{1}{3} \right) \)
• B. \( \frac{1}{3} \geq P(A \cap B) \geq \frac{1}{12} \)
• C. \( \frac{3}{4} \geq P(A \cap B) \geq \frac{1}{3} \)
• D. \( P(A \cap B) \geq \min\left( \frac{3}{4}, \frac{1}{3} \right) \)

Hint:

The probability of the intersection of two events is always between the maximum of their individual probabilities and the product of the probabilities.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between events.
For two events A and B, the probability of their intersection \( P(A \cap B) \) is bounded by:
\[ P(A \cap B) \leq \min(P(A), P(B)). \] Thus, \( P(A \cap B) \leq \min\left( \frac{3}{4}, \frac{1}{3} \right) = \frac{1}{3} \). This means \( P(A \cap B) \) cannot exceed \( \frac{1}{3} \).
Step 2: Lower bound for the intersection.
The probability of the intersection is also bounded below by the product of the individual probabilities: \[ P(A \cap B) \geq P(A) + P(B) - 1 = \frac{3}{4} + \frac{1}{3} - 1 = \frac{9}{12} + \frac{4}{12} - \frac{12}{12} = \frac{1}{12}. \] Step 3: Conclusion.
Thus, \( P(A \cap B) \) is between \( \frac{1}{12} \) and \( \frac{1}{3} \), which corresponds to option (B). Final Answer: (B) \( \frac{1}{3} \geq P(A \cap B) \geq \frac{1}{12} \)