Suppose that

Box-I contains 8 red, 3 blue and 5 green balls,

Box-II contains 24 red, 9 blue and 15 green balls,

Box-III contains 1 blue, 12 green and 3 yellow balls,

Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $$b$$. If $$b$$ is red then a ball is chosen randomly from Box-II, if $$b$$ is blue then a ball is chosen randomly from Box-III, and if $$b$$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

Let the two events be defined as follows:
  • $$A$$ : “one of the two chosen balls is white’’
  • $$B$$ : “at least one of the two chosen balls is green’’

We must evaluate the conditional probability $$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$$.

Step 1 : Probability of each colour chosen from Box I
Box I has 8 red, 3 blue and 5 green balls; total $$16$$ balls.

$$P(R_1)=\frac{8}{16}=\frac12,\quad P(B_1)=\frac{3}{16},\quad P(G_1)=\frac{5}{16}$$

(Here the subscript ‘1’ marks the first ball.)

Step 2 : Composition of the other boxes
Box II : 24 R, 9 B, 15 G  (total 48)
Box III: 1 B, 12 G, 3 Y (total 16) (no white)
Box IV : 10 G, 16 O, 6 W (total 32)

Step 3 : Compute $$P(A\cap B)$$
The event $$A\cap B$$ requires that a white ball is obtained and at least one of the two balls is green.

Case 1 : first ball red
Probability $$\dfrac12$$. The second ball then comes from Box II, which contains no white balls, so $$A$$ cannot occur. Hence this case contributes 0 to $$P(A\cap B)$$.

Case 2 : first ball blue
Probability $$\dfrac{3}{16}$$. The second ball comes from Box III, which also has no white balls, so again $$A$$ is impossible. Contribution = 0.

Case 3 : first ball green
Probability $$\dfrac{5}{16}$$. Here $$B$$ is already satisfied because the first ball is green. The second ball is drawn from Box IV, which contains 6 white balls out of 32.
$$P(\text{second ball white}\mid G_1)=\frac{6}{32}=\frac{3}{16}$$

Therefore the probability of both $$A$$ and $$B$$ occurring is
$$P(A\cap B)=\frac{5}{16}\times\frac{3}{16}=\frac{15}{256}$$

Step 4 : Compute $$P(B)$$

Case 1 : first ball red
We need the second ball (from Box II) to be green.
$$P(G_2\mid R_1)=\frac{15}{48}=\frac{5}{16}$$
Contribution: $$\frac12\times\frac{5}{16}=\frac{5}{32}$$

Case 2 : first ball blue
Second ball is from Box III; it must be green.
$$P(G_2\mid B_1)=\frac{12}{16}=\frac34$$
Contribution: $$\frac{3}{16}\times\frac34=\frac{9}{64}$$

Case 3 : first ball green
Event $$B$$ is automatically satisfied regardless of the second draw.
Contribution: $$\frac{5}{16}\times 1=\frac{5}{16}$$

Add the three contributions:
$$P(B)=\frac{5}{32}+\frac{9}{64}+\frac{5}{16}$$

Convert to a common denominator 64:
$$P(B)=\frac{10}{64}+\frac{9}{64}+\frac{20}{64}=\frac{39}{64}$$

Step 5 : Conditional probability
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}= \frac{\dfrac{15}{256}}{\dfrac{39}{64}}= \frac{15}{256}\times\frac{64}{39}$$

Simplify: $$\frac{64}{256}=\frac14\;,\quad P(A\mid B)=\frac{15}{4\times39}=\frac{15}{156}=\frac{5}{52}$$

Hence the required conditional probability is $$\dfrac{5}{52}$$.

Option C which is: $$\dfrac{5}{52}$$.