Conditional Probability of Drawing White Ball Given Green Ball

To find the required conditional probability, let us first define the composition of each box and the events involved in the experiment.

1. Composition of the Boxes

• Box-I: $8$ Red ($R$), $3$ Blue ($B$), $5$ Green ($G$).
  $\text{Total} = 8 + 3 + 5 = 16$ balls.
• Box-II: $24$ Red ($R$), $9$ Blue ($B$), $15$ Green ($G$).
  $\text{Total} = 24 + 9 + 15 = 48$ balls.
• Box-III: $1$ Blue ($B$), $12$ Green ($G$), $3$ Yellow ($Y$).
  $\text{Total} = 1 + 12 + 3 = 16$ balls.
• Box-IV: $10$ Green ($G$), $16$ Orange ($O$), $6$ White ($W$).
  $\text{Total} = 10 + 16 + 6 = 32$ balls.

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2. Define the Events

Let:

• $R_1, B_1, G_1$ be the events of drawing a Red, Blue, or Green ball from Box-I, respectively.
• $E$ be the event that at least one of the chosen balls is Green.
• $W$ be the event that one of the chosen balls is White.

We are required to compute the conditional probability $P(W \mid E)$, given by: $P(W \mid E) = \frac{P(W \cap E)}{P(E)}$

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3. Calculate $P(E)$ (Probability of at least one Green ball)

There are three mutually exclusive cases based on the first ball chosen from Box-I:

1. Case 1: First ball chosen from Box-I is Red ($R_1$) $P(R_1) = \frac{8}{16} = \frac{1}{2}$ Then a ball is drawn from Box-II. For at least one green ball to occur, the ball drawn from Box-II must be Green ($G_2$): $P(G_2 \mid R_1) = \frac{15}{48} = \frac{5}{16}$ $P(R_1 \cap G_2) = \frac{1}{2} \times \frac{5}{16} = \frac{5}{32} = \frac{40}{256}$

2. Case 2: First ball chosen from Box-I is Blue ($B_1$) $P(B_1) = \frac{3}{16}$ Then a ball is drawn from Box-III. For at least one green ball to occur, the ball drawn from Box-III must be Green ($G_3$): $P(G_3 \mid B_1) = \frac{12}{16} = \frac{3}{4}$ $P(B_1 \cap G_3) = \frac{3}{16} \times \frac{3}{4} = \frac{9}{64} = \frac{36}{256}$

3. Case 3: First ball chosen from Box-I is Green ($G_1$) $P(G_1) = \frac{5}{16} = \frac{80}{256}$ Since the first ball itself is Green, event $E$ occurs regardless of which ball is subsequently drawn from Box-IV. Thus: $P(G_1 \cap E) = P(G_1) = \frac{80}{256}$

Summing the probabilities for these mutually exclusive cases gives: $P(E) = P(R_1 \cap G_2) + P(B_1 \cap G_3) + P(G_1)$ $P(E) = \frac{40}{256} + \frac{36}{256} + \frac{80}{256} = \frac{156}{256}$

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4. Calculate $P(W \cap E)$

White balls are present only in Box-IV, which is accessed if and only if the ball drawn from Box-I is Green ($G_1$).

Since drawing a White ball requires $G_1$ to be chosen first, any outcome containing a White ball already contains at least one Green ball ($G_1$). Hence, $W \subseteq E$, which implies: $P(W \cap E) = P(W)$

Now, the probability of drawing a White ball $W$ is: $P(W) = P(G_1) \times P(W_4 \mid G_1)$ $P(W) = \frac{5}{16} \times \frac{6}{32} = \frac{30}{512} = \frac{15}{256}$

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5. Calculate the Conditional Probability $P(W \mid E)$

$P(W \mid E) = \frac{P(W \cap E)}{P(E)} = \frac{P(W)}{P(E)} = \frac{\frac{15}{256}}{\frac{156}{256}} = \frac{15}{156}$

Simplifying the fraction by dividing both numerator and denominator by $3$: $P(W \mid E) = \frac{5}{52}$

Thus, the correct option is (C).