Conditional Probability

We need to find $$ P(\text{first black} \mid \text{second black}) $$

We know, for conditional probability, $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$

Let, $$ A$$ be the event that the first ball is black, and $$ B$$ be that the second ball is black

So, probability both balls are black (without replacement) is $$ P(A \cap B) = \dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{1}{3} $$

Probability second ball is black is 

$$ P(B) = P(\text{second black | first black}) \cdot P(\text{first black}) + P(\text{second black | first white}) \cdot P(\text{first white}) $$

$$ = \dfrac{5}{9} \cdot \dfrac{6}{10} + \dfrac{6}{9} \cdot \dfrac{4}{10} = \dfrac{30}{90} + \dfrac{24}{90} = \dfrac{3}{5} $$

Now,  $$ P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/3}{3/5} = \dfrac{5}{9} $$

Thus, $$ m+n = 5+9 =14$$ .

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.

To find this conditional probability efficiently, we use modulo 3 arithmetic to count the valid pairs in our reduced sample space.

Step 1: Categorize the elements of the set modulo 3

The set $$\{1, 2, 3, \dots, 30\}$$ contains 30 elements. We can divide them into three categories based on their remainders when divided by 3:

Category R0 (Remainder 0): $$\{3, 6, 9, \dots, 30\}$$ containing 10 numbers.

Category R1 (Remainder 1): $$\{1, 4, 7, \dots, 28\}$$ containing 10 numbers.

Category R2 (Remainder 2): $$\{2, 5, 8, \dots, 29\}$$ containing 10 numbers.

Step 2: Calculate the size of the reduced sample space

We are given that $$a^2 - b^2$$ is divisible by 3. Factoring the algebraic expression gives: $$a^2 - b^2 = (a - b)(a + b)$$

For a product to be divisible by the prime number 3, either $$(a - b)$$ must be divisible by 3, or $$(a + b)$$ must be divisible by 3 (or both). Let's count the total pairs $$(a, b)$$ where this happens:

Case 1: $$(a - b)$$ is divisible by 3

This occurs when both $$a$$ and $$b$$ have the same remainder when divided by 3:

If both belong to R0: $$10 \times 9 = 90$$ ordered pairs.

If both belong to R1: $$10 \times 9 = 90$$ ordered pairs.

If both belong to R2: $$10 \times 9 = 90$$ ordered pairs.

$$\text{Total for Case 1} = 90 + 90 + 90 = 270 \text{ pairs}$$

Case 2: $$(a + b)$$ is divisible by 3

This occurs when the sum of their remainders is a multiple of 3. This means either both belong to R0 (already completely counted in Case 1), or one number belongs to R1 and the other belongs to R2:

One from R1 and one from R2: $$10 \times 10 \times 2 = 200$$ ordered pairs.

Adding these two independent possibilities gives the total sample space size:

$$\text{Total Valid Pairs} = 270 + 200 = 470 \text{ pairs}$$

Step 3: Count the favorable pairs

We need to find the number of pairs where $$a + b$$ is divisible by 3 out of our valid 470 pairs. From our analysis above, $$a + b$$ is divisible by 3 in:

Pairs where both belong to R0: 90 pairs

Pairs where one belongs to R1 and the other belongs to R2: 200 pairs

$$\text{Favorable Pairs} = 90 + 200 = 290 \text{ pairs}$$ 

Step 4: Calculate the probability

$$\text{Probability} = \frac{\text{Favorable Pairs}}{\text{Total Valid Pairs}}$$

$$\text{Probability} = \frac{290}{470} = \frac{29}{47}$$