Suppose A and B are two events such that P(A|B)=0.6, P(B|A)=0.3 , P(A)=0.1 then `P(A' nn B')` is equal to _____

Answer

To solve the problem, we need to find \( P(A' \cap B') \) given the probabilities \( P(A|B) = 0.6 \), \( P(B|A) = 0.3 \), and \( P(A) = 0.1 \).

### Step-by-Step Solution:

1. **Use the definition of conditional probability**:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
From this, we can express \( P(A \cap B) \):
\[ P(A \cap B) = P(A|B) \cdot P(B) = 0.6 \cdot P(B) \]

2. **Similarly, use the definition of conditional probability for \( P(B|A) \)**:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
Thus, we can express \( P(A \cap B) \) in another way:
\[ P(A \cap B) = P(B|A) \cdot P(A) = 0.3 \cdot 0.1 = 0.03 \]

3. **Set the two expressions for \( P(A \cap B) \) equal to each other**:
\[ 0.6 \cdot P(B) = 0.03 \]
Solving for \( P(B) \):
\[ P(B) = \frac{0.03}{0.6} = 0.05 \]

4. **Now, we have \( P(A) \) and \( P(B) \)**:
- \( P(A) = 0.1 \)
- \( P(B) = 0.05 \)

5. **Use the formula for \( P(A \cup B) \)**:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ P(A \cup B) = 0.1 + 0.05 - 0.03 = 0.12 \]

6. **Now, find \( P(A' \cap B') \)** using De Morgan's Law:
\[ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \]
Thus,
\[ P(A' \cap B') = 1 - 0.12 = 0.88 \]

### Final Answer:
\[ P(A' \cap B') = 0.88 \]