Question

Let \( E, F \), and \( G \) be any three events with \( P(E) = 0.3 \), \( P(F|E) = 0.2 \), \( P(G|E) = 0.1 \). Then \( P(E - (F \cup G)) \) equals

• A. 0.155
• B. 0.175
• C. 0.225
• D. 0.255

Hint:

For conditional probabilities, remember that \( P(A \cap B) = P(B|A) \cdot P(A) \).

Answer 1

The Correct Option is C

Step 1: Use the formula for conditional probability.
The formula for \( P(E - (F \cup G)) \) is: \[ P(E - (F \cup G)) = P(E) - P(E \cap (F \cup G)) = P(E) - P(E \cap F) - P(E \cap G) \] Step 2: Substitute the given values.
We are given \( P(E) = 0.3 \), \( P(F|E) = 0.2 \), and \( P(G|E) = 0.1 \), so we can calculate: \[ P(E \cap F) = P(F|E) \cdot P(E) = 0.2 \cdot 0.3 = 0.06 \] \[ P(E \cap G) = P(G|E) \cdot P(E) = 0.1 \cdot 0.3 = 0.03 \] Step 3: Conclusion.
Thus, \( P(E - (F \cup G)) = 0.3 - 0.06 - 0.03 = 0.175 \). The correct answer is (B) 0.175.