Let A and E be any two events with positive probabilities.
Statement I: $$P(E/A) \geq P(A/E)P(E)$$.
Statement II: $$P(A/E) \geq P(A \cap E)$$.

Let us solve the given problem step by step. We have two events, A and E, both with positive probabilities. We need to evaluate two statements.

First, recall the definition of conditional probability. The conditional probability of E given A is defined as:

$$P(E|A) = \frac{P(A \cap E)}{P(A)}$$

provided that $$P(A) > 0$$. Similarly, the conditional probability of A given E is:

$$P(A|E) = \frac{P(A \cap E)}{P(E)}$$

provided that $$P(E) > 0$$. Since both events have positive probabilities, these definitions are valid.

Now, let us analyze Statement I: $$P(E|A) \geq P(A|E)P(E)$$.

Substitute the expressions for the conditional probabilities:

Left side: $$P(E|A) = \frac{P(A \cap E)}{P(A)}$$

Right side: $$P(A|E)P(E) = \left( \frac{P(A \cap E)}{P(E)} \right) \times P(E) = P(A \cap E)$$

So, Statement I becomes:

$$\frac{P(A \cap E)}{P(A)} \geq P(A \cap E)$$

Let $$x = P(A \cap E)$$ and $$a = P(A)$$. Since probabilities are non-negative, $$x \geq 0$$, and given that $$P(A) > 0$$, $$a > 0$$. Also, since $$P(A) \leq 1$$, $$a \leq 1$$. The inequality is:

$$\frac{x}{a} \geq x$$

We can rearrange this as:

$$\frac{x}{a} - x \geq 0$$

$$x \left( \frac{1}{a} - 1 \right) \geq 0$$

Since $$a \leq 1$$, $$\frac{1}{a} \geq 1$$, so $$\frac{1}{a} - 1 \geq 0$$. Also, $$x \geq 0$$. Therefore, the product $$x \left( \frac{1}{a} - 1 \right)$$ is non-negative. This holds for all values of $$x$$ and $$a$$:

• If $$x = 0$$, both sides are 0, so $$0 \geq 0$$ is true.
• If $$x > 0$$, since $$\frac{1}{a} - 1 \geq 0$$, the inequality holds.

Thus, Statement I is always true.

Now, consider Statement II: $$P(A|E) \geq P(A \cap E)$$.

Substitute the expression for $$P(A|E)$$:

$$P(A|E) = \frac{P(A \cap E)}{P(E)}$$

So, Statement II becomes:

$$\frac{P(A \cap E)}{P(E)} \geq P(A \cap E)$$

Let $$x = P(A \cap E)$$ and $$e = P(E)$$. Since $$P(E) > 0$$, $$e > 0$$, and $$P(E) \leq 1$$, so $$e \leq 1$$. The inequality is:

$$\frac{x}{e} \geq x$$

Rearrange this as:

$$\frac{x}{e} - x \geq 0$$

$$x \left( \frac{1}{e} - 1 \right) \geq 0$$

Since $$e \leq 1$$, $$\frac{1}{e} \geq 1$$, so $$\frac{1}{e} - 1 \geq 0$$. Also, $$x \geq 0$$. Therefore, the product $$x \left( \frac{1}{e} - 1 \right)$$ is non-negative. This holds for all values of $$x$$ and $$e$$:

• If $$x = 0$$, both sides are 0, so $$0 \geq 0$$ is true.
• If $$x > 0$$, since $$\frac{1}{e} - 1 \geq 0$$, the inequality holds.

Thus, Statement II is also always true.

Since both statements are true, the correct option is B.

Hence, the correct answer is Option B.