Question

If \( P(A) = 0.12, P(B) = 0.15 \) and \( P(B/A) = 0.18 \), then find the value of \( P(A \cap B) \).

Hint:

{Formula:} \[ P(B/A) = \frac{P(A \cap B)}{P(A)} \quad \Rightarrow \quad P(A \cap B) = P(A) \times P(B/A) \] Always ensure probabilities are in decimal form before multiplication.

Answer 1

Step 1: Recall the formula for conditional probability \[ P(B/A) = \frac{P(A \cap B)}{P(A)} \]
Step 2: Substitute the given values Given: \[ P(A) = 0.12, \quad P(B/A) = 0.18 \] \[ 0.18 = \frac{P(A \cap B)}{0.12} \]
Step 3: Solve for \( P(A \cap B) \) \[ P(A \cap B) = 0.18 \times 0.12 \] \[ P(A \cap B) = 0.0216 \]
Step 4: Final answer \[ \boxed{0.0216} \] Verification: We can verify using the multiplication rule: \[ P(A \cap B) = P(A) \times P(B/A) = 0.12 \times 0.18 = 0.0216 \]