Step-by-step Solution

1. Define the experimental setup and initial probabilities.

Let's denote the contents of the boxes and the probabilities of choosing a ball of a certain color from Box-I.

• Box-I: 8 Red (R1), 3 Blue (B1), 5 Green (G1). Total = 16 balls.

  • $P(R_1) = \frac{8}{16} = \frac{1}{2}$
  • $P(B_1) = \frac{3}{16}$
  • $P(G_1) = \frac{5}{16}$

• Box-II (if R1 is chosen): 24 Red (R2), 9 Blue (B2), 15 Green (G2). Total = 48 balls.

• Box-III (if B1 is chosen): 1 Blue (B3), 12 Green (G3), 3 Yellow (Y3). Total = 16 balls.

• Box-IV (if G1 is chosen): 10 Green (G4), 16 Orange (O4), 6 White (W4). Total = 32 balls.

2. Define the events for conditional probability.

We need to find the conditional probability $P(E|F)$, where:

• $E$ = The event that 'one of the chosen balls is white'.
• $F$ = The event that 'at least one of the chosen balls is green'.

The formula for conditional probability is $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

3. Calculate the probability of event F, $P(F)$.

Event $F$ ('at least one green ball') can happen in the following mutually exclusive ways:

• Case 1: The first ball from Box-I is green ($G_1$). The second ball can be anything. The condition is already met.

  • The probability of this case is $P(G_1) = \frac{5}{16}$.

• Case 2: The first ball is red ($R_1$) and the second ball is green ($G_2$).

  • $P(R_1 \cap G_2) = P(R_1) \times P(G_2 | R_1) = \frac{8}{16} \times \frac{15}{48} = \frac{1}{2} \times \frac{5}{16} = \frac{5}{32}$.

• Case 3: The first ball is blue ($B_1$) and the second ball is green ($G_3$).

  • $P(B_1 \cap G_3) = P(B_1) \times P(G_3 | B_1) = \frac{3}{16} \times \frac{12}{16} = \frac{3}{16} \times \frac{3}{4} = \frac{9}{64}$.

The total probability of event $F$ is the sum of the probabilities of these cases: $P(F) = P(G_1) + P(R_1 \cap G_2) + P(B_1 \cap G_3)$ $P(F) = \frac{5}{16} + \frac{5}{32} + \frac{9}{64}$ To add these fractions, we find a common denominator, which is 64: $P(F) = \frac{5 \times 4}{16 \times 4} + \frac{5 \times 2}{32 \times 2} + \frac{9}{64} = \frac{20}{64} + \frac{10}{64} + \frac{9}{64} = \frac{39}{64}$

4. Calculate the probability of the intersection event, $P(E \cap F)$.

• Event $E$ ('one white ball') can only occur if the second ball is white. White balls are only in Box-IV. To choose from Box-IV, the first ball must be green ($G_1$).
• So, event $E$ is the sequence of drawing a green ball from Box-I and then a white ball from Box-IV. $P(E) = P(G_1 \cap W_4) = P(G_1) \times P(W_4 | G_1) = \frac{5}{16} \times \frac{6}{32} = \frac{5}{16} \times \frac{3}{16} = \frac{15}{256}$
• Now consider the intersection event $E \cap F$: 'one white ball' AND 'at least one green ball'.
• If event $E$ occurs, the first ball drawn is green ($G_1$). This means the condition for event $F$ ('at least one green ball') is automatically satisfied.
• Therefore, event $E$ is a subset of event $F$ ($E \subset F$), which implies that $E \cap F = E$.
• So, $P(E \cap F) = P(E) = \frac{15}{256}$.

5. Calculate the conditional probability $P(E|F)$.

Using the formula for conditional probability: $P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{15/256}{39/64}$ $P(E|F) = \frac{15}{256} \times \frac{64}{39}$ Since $256 = 4 \times 64$, we can simplify: $P(E|F) = \frac{15}{4 \times 39}$ Now, we can simplify the fraction by dividing the numerator and denominator by 3: $P(E|F) = \frac{3 \times 5}{4 \times (3 \times 13)} = \frac{5}{4 \times 13} = \frac{5}{52}$

Thus, the required conditional probability is $\frac{5}{52}$.

Conclusion: Comparing the result with the given options, the correct option is C.