Question

If A and B are events such that \( P(A) = 0.3, P(B) = 0.2 \) and \( P(A \cup B) = 0.45 \), the value of \( P(A \cap B) \) is .............

Hint:

To find the intersection of two events, use the formula \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \).

Answer 1

Correct Answer: 0.24 - 0.26

Step 1: Use the addition rule

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

$$0.45 = 0.3 + 0.2 - P(A \cap B)$$

$$0.45 = 0.5 - P(A \cap B)$$

$$P(A \cap B) = 0.5 - 0.45 = 0.05$$

Step 2: Partition event A

Event $A$ can be partitioned into two mutually exclusive parts:

• $A \cap B$ (A occurs and B occurs)
• $A \cap \bar{B}$ (A occurs and B does not occur)

Therefore: $$P(A) = P(A \cap B) + P(A \cap \bar{B})$$

Step 3: Solve for $P(A \cap \bar{B})$

$$P(A \cap \bar{B}) = P(A) - P(A \cap B)$$

$$= 0.3 - 0.05$$

$$= 0.25$$

Answer: 0.25