Independent Events

Let us determine the total counts of the balls and find the probabilities for each event.

Box I contains:

$$ \text{Red balls} = 6 $$

$$ \text{Green balls} = 9 $$

$$ \text{Total balls in Box I} = 6 + 9 = 15 $$

Box II contains:

$$ \text{Red balls} = 8 $$

$$ \text{Green balls} = 12 $$

$$ \text{Total balls in Box II} = 8 + 12 = 20 $$

All balls are mixed together:

$$ \text{Total Red balls } (F_1) = 6 + 8 = 14 $$

$$ \text{Total Green balls } (F_2) = 9 + 12 = 21 $$

$$ \text{Overall Total balls} = 15 + 20 = 35 $$

Step 1: Calculate Individual Probabilities

The probability that a chosen ball belongs to Box I is:

$$ P(E_1) = \frac{15}{35} = \frac{3}{7} $$

The probability that a chosen ball belongs to Box II is:

$$ P(E_2) = \frac{20}{35} = \frac{4}{7} $$

The probability that a chosen ball is red is:

$$ P(F_1) = \frac{14}{35} = \frac{2}{5} $$

The probability that a chosen ball is green is:

$$ P(F_2) = \frac{21}{35} = \frac{3}{5} $$

Step 2: Evaluate Independence of $$ E_1 $$ and $$ F_1 $$

The intersection probability $$ P(E_1 \cap F_1) $$ represents choosing a red ball that came from Box I:

$$ P(E_1 \cap F_1) = \frac{6}{35} $$

Now, let us calculate the product of their individual probabilities:

$$ P(E_1) \cdot P(F_1) = \frac{3}{7} \cdot \frac{2}{5} = \frac{6}{35} $$

Since $$ P(E_1 \cap F_1) = P(E_1) \cdot P(F_1) $$, the events $$ E_1 $$ and $$ F_1 $$ are independent. Therefore, Statement A is TRUE.

Step 3: Evaluate Independence of $$ E_2 $$ and $$ F_2 $$

The intersection probability $$ P(E_2 \cap F_2) $$ represents choosing a green ball that came from Box II:

$$ P(E_2 \cap F_2) = \frac{12}{35} $$

Now, let us calculate the product of their individual probabilities:

$$ P(E_2) \cdot P(F_2) = \frac{4}{7} \cdot \frac{3}{5} = \frac{12}{35} $$

Since $$ P(E_2 \cap F_2) = P(E_2) \cdot P(F_2) $$, the events $$ E_2 $$ and $$ F_2 $$ are independent. This means any statement claiming they are dependent is FALSE.

Step 4: Evaluate Conditional Probabilities

Let us calculate $$ P(F_1 \mid E_1) $$:

$$ P(F_1 \mid E_1) = \frac{P(E_1 \cap F_1)}{P(E_1)} = \frac{\frac{6}{35}}{\frac{15}{35}} = \frac{6}{15} = \frac{2}{5} $$

Let us calculate $$ P(F_1 \mid E_2) $$:

$$ P(F_1 \mid E_2) = \frac{P(E_2 \cap F_1)}{P(E_2)} = \frac{\frac{8}{35}}{\frac{20}{35}} = \frac{8}{20} = \frac{2}{5} $$

Since $$ P(F_1 \mid E_1) = P(F_1 \mid E_2) = \frac{2}{5} $$, the conditional probabilities are equal. Therefore, Statement C is FALSE.

Let us check the last comparison involving $$ P(F_2 \mid E_2) $$:

$$ P(F_2 \mid E_2) = \frac{P(E_2 \cap F_2)}{P(E_2)} = \frac{\frac{12}{35}}{\frac{20}{35}} = \frac{12}{20} = \frac{3}{5} $$

Comparing the values, $$ \frac{2}{5} $$ is not greater than $$ \frac{3}{5} $$, making any statement asserting $$ P(F_1 \mid E_1) > P(F_2 \mid E_2) $$ FALSE.

Conclusion

The correct statements are:

A. The events $$ E_1 $$ and $$ F_1 $$ are independent.