Probability of Independent and Conditional Events in Two Boxes

To determine the correct options, we analyze the total composition of the boxes and calculate the probabilities of the given events.

1. Composition of the Boxes:

• Box I: Contains $6$ red balls and $9$ green balls.
  • Total number of balls in Box I = $6 + 9 = 15$.
• Box II: Contains $8$ red balls and $12$ green balls.
  • Total number of balls in Box II = $8 + 12 = 20$.

When all the balls from Box I and Box II are mixed together, the combined total number of balls is: $\text{Total balls} = 15 + 20 = 35$

The total number of red balls and green balls in the combined mix are: $\text{Total Red balls} = 6 + 8 = 14$ $\text{Total Green balls} = 9 + 12 = 21$

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2. Probabilities of Individual Events:

A ball is chosen at random from the total $35$ mixed balls.

• Event $E_1$ (Ball belongs to Box I): $P(E_1) = \frac{15}{35} = \frac{3}{7}$

• Event $E_2$ (Ball belongs to Box II): $P(E_2) = \frac{20}{35} = \frac{4}{7}$

• Event $F_1$ (Ball is red): $P(F_1) = \frac{14}{35} = \frac{2}{5}$

• Event $F_2$ (Ball is green): $P(F_2) = \frac{21}{35} = \frac{3}{5}$

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3. Verification of Options:

Option A:

• Probability that the ball is chosen from Box I and is red: $P(E_1 \cap F_1) = \frac{\text{Red balls in Box I}}{\text{Total combined balls}} = \frac{6}{35}$
• Product of individual probabilities: $P(E_1) \cdot P(F_1) = \frac{3}{7} \times \frac{2}{5} = \frac{6}{35}$

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

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Option B:

• Probability that the ball is chosen from Box II and is green: $P(E_2 \cap F_2) = \frac{\text{Green balls in Box II}}{\text{Total combined balls}} = \frac{12}{35}$
• Product of individual probabilities: $P(E_2) \cdot P(F_2) = \frac{4}{7} \times \frac{3}{5} = \frac{12}{35}$

Since $P(E_2 \cap F_2) = P(E_2) \cdot P(F_2)$, the events $E_2$ and $F_2$ are independent (not dependent).
Thus, Option B is FALSE.

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Option C:

• Conditional probability $P(F_1 \mid E_1)$: $P(F_1 \mid E_1) = \frac{\text{Red balls in Box I}}{\text{Total balls in Box I}} = \frac{6}{15} = \frac{2}{5}$
• Conditional probability $P(F_1 \mid E_2)$: $P(F_1 \mid E_2) = \frac{\text{Red balls in Box II}}{\text{Total balls in Box II}} = \frac{8}{20} = \frac{2}{5}$

Since $P(F_1 \mid E_1) = P(F_1 \mid E_2) = \frac{2}{5}$, the two conditional probabilities are equal.
Thus, Option C is TRUE.

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Option D:

• From above, $P(F_1 \mid E_1) = \frac{2}{5}$.
• Conditional probability $P(F_2 \mid E_2)$: $P(F_2 \mid E_2) = \frac{\text{Green balls in Box II}}{\text{Total balls in Box II}} = \frac{12}{20} = \frac{3}{5}$

Comparing the two: $P(F_1 \mid E_1) = \frac{2}{5} < \frac{3}{5} = P(F_2 \mid E_2)$

Therefore, $P(F_1 \mid E_1)$ is less than $P(F_2 \mid E_2)$.
Thus, Option D is FALSE.

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Conclusion:

The correct options are A and C.