Statement-1: if A and B are two events, such that `0 lt P(A),P(B) lt 1,` then `P((A)/(overline(B)))+P((overline(A))/(overline(B)))=(3)/(2)`
Statement-2: If A and B are two events, such that `0 lt P(A), P(B) lt 1,` then
`P(A//B)=(P(AcapB))/(P(B)) and P(overline(B))=P(A capoverline(B))+P(overline(A) cap overline(B))`

Options

A. Statement-1 is true, Statement-2 is true: Statement-2 is a correct explanation for Statement-1
B. Statement-1 is true, Statement-2 is true: Statement-2 is not a correct explanation for Statement-1
C. Statement-1 is true, Statement-2 is false
D. Statement-1 is false, Statement-2 is true

Answer

To solve the problem, we need to evaluate the two statements provided and determine their validity.

### Step-by-Step Solution:

**Step 1: Analyze Statement 1**
- Statement 1 claims that if \( A \) and \( B \) are two events such that \( 0 < P(A), P(B) < 1 \), then:
\[ P(A | \overline{B}) + P(\overline{A} | \overline{B}) = \frac{3}{2} \]

**Step 2: Use the Definition of Conditional Probability**
- Recall that the conditional probability is defined as:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
- Therefore, we can express \( P(A | \overline{B}) \) and \( P(\overline{A} | \overline{B}) \) as:
\[ P(A | \overline{B}) = \frac{P(A \cap \overline{B})}{P(\overline{B})} \]
\[ P(\overline{A} | \overline{B}) = \frac{P(\overline{A} \cap \overline{B})}{P(\overline{B})} \]

**Step 3: Combine the Expressions**
- Combining these two expressions gives:
\[ P(A | \overline{B}) + P(\overline{A} | \overline{B}) = \frac{P(A \cap \overline{B}) + P(\overline{A} \cap \overline{B})}{P(\overline{B})} \]

**Step 4: Simplify the Numerator**
- The numerator can be simplified:
\[ P(A \cap \overline{B}) + P(\overline{A} \cap \overline{B}) = P(\overline{B}) \quad \text{(since these two events cover all outcomes in } \overline{B}\text{)} \]

**Step 5: Substitute Back**
- Therefore, we have:
\[ P(A | \overline{B}) + P(\overline{A} | \overline{B}) = \frac{P(\overline{B})}{P(\overline{B})} = 1 \]

**Conclusion for Statement 1:**
- Since we found that \( P(A | \overline{B}) + P(\overline{A} | \overline{B}) = 1 \), Statement 1 is **false** because it claims the sum equals \( \frac{3}{2} \).

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**Step 6: Analyze Statement 2**
- Statement 2 states:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \quad \text{and} \quad P(\overline{B}) = P(A \cap \overline{B}) + P(\overline{A} \cap \overline{B}) \]

**Step 7: Verify Statement 2**
- The first part is the definition of conditional probability and is always true.
- The second part states that the probability of \( \overline{B} \) can be expressed as the sum of the probabilities of the intersections with \( A \) and \( \overline{A} \), which is also true by the law of total probability.

**Conclusion for Statement 2:**
- Both parts of Statement 2 are true, thus Statement 2 is **true**.

### Final Answer:
- **Statement 1** is **false**.
- **Statement 2** is **true**.

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