Question

The events A and B are such that \(P(A)=0.7, P(B')=0.8\) and \(P(A \cup B) = 0.8\). Then \(P(A' \cup B')\) is equal to

• A. 0.9
• B. 0.7
• C. 0.5
• D. 0.3
• E. 0.1

Hint:

Always keep De Morgan's laws handy: \(P(A' \cup B') = 1 - P(A \cap B)\) and \(P(A' \cap B') = 1 - P(A \cup B)\). These identities solve 90% of intersection/union probability problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the addition theorem of probability and De Morgan's laws to find the required probability.

Step 2: Key Formula or Approach:
1. \(P(B) = 1 - P(B')\).
2. \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\).
3. De Morgan's Law: \(A' \cup B' = (A \cap B)'\).
4. \(P((A \cap B)') = 1 - P(A \cap B)\).

Step 3: Detailed Explanation:
1. Find \(P(B)\): \[ P(B) = 1 - 0.8 = 0.2 \]
2. Find \(P(A \cap B)\): \[ P(A \cap B) = 0.7 + 0.2 - 0.8 = 0.1 \]
3. Find \(P(A' \cup B')\): \[ P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) = 1 - 0.1 = 0.9 \]

Step 4: Final Answer:
The probability is 0.9.