Let A and E be two events with positive probabilites
Statement 1 : `P(E//A)ge P(A//E) P(E)`
Statement 2 : `P(A//E)ge P(A cap E)`

Options

A. both statements are true
B. both statements are false
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 analyze the two statements given about the events A and E with positive probabilities.

### Step 1: Understand the Definitions
We start with the definitions of conditional probability:
- The conditional probability of E given A is defined as:
\[ P(E | A) = \frac{P(E \cap A)}{P(A)} \]
- The conditional probability of A given E is defined as:
\[ P(A | E) = \frac{P(A \cap E)}{P(E)} \]

### Step 2: Write the Equations
From the definitions, we can write:
1. \( P(E | A) = \frac{P(E \cap A)}{P(A)} \) (Equation 1)
2. \( P(A | E) = \frac{P(A \cap E)}{P(E)} \) (Equation 2)

### Step 3: Divide the Two Equations
Now, we divide Equation 1 by Equation 2:
\[ \frac{P(E | A)}{P(A | E)} = \frac{\frac{P(E \cap A)}{P(A)}}{\frac{P(A \cap E)}{P(E)}} \]
This simplifies to:
\[ \frac{P(E | A)}{P(A | E)} = \frac{P(E \cap A) \cdot P(E)}{P(A \cap E) \cdot P(A)} \]
Since \( P(E \cap A) = P(A \cap E) \), we can cancel these terms:
\[ \frac{P(E | A)}{P(A | E)} = \frac{P(E)}{P(A)} \]

### Step 4: Analyze Statement 1
From the equation derived, we can express:
\[ P(E | A) \geq P(A | E) \cdot P(E) \]
This means that Statement 1 \( P(E | A) \geq P(A | E) \cdot P(E) \) is true.

### Step 5: Analyze Statement 2
Now, we analyze Statement 2:
\[ P(A | E) \geq P(A \cap E) \]
Using the definition of conditional probability:
\[ P(A | E) = \frac{P(A \cap E)}{P(E)} \]
Since \( P(E) > 0 \) (as given), we can multiply both sides by \( P(E) \):
\[ P(A | E) \cdot P(E) \geq P(A \cap E) \]
This implies:
\[ P(A | E) \geq P(A \cap E) \]
Thus, Statement 2 is also true.

### Conclusion
Both statements are true:
1. Statement 1: True
2. Statement 2: True