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

The correct answer is $\frac{5}{52}$.

We are asked to find the conditional probability of the event W = "one of the chosen balls is white" given the event G = "at least one of the chosen balls is green".

That is, we need: $\frac{P(W\cap G)}{P\left(G\right)}$

Step 1: Understand the setup.

Box-I has 8 red + 3 blue + 5 green = 16 balls total.
Ball b is drawn from Box-I. Then:

- If b is red → draw from Box-II (24 red, 9 blue, 15 green → 48 balls total)
- If b is blue → draw from Box-III (1 blue, 12 green, 3 yellow → 16 balls total)
- If b is green → draw from Box-IV (10 green, 16 orange, 6 white → 32 balls total)

Step 2: Identify the probabilities of choosing each color from Box-I.

$P\left(\text{red from Box-I}\right)=\frac{8}{16}=\frac{1}{2}$

$P\left(\text{blue from Box-I}\right)=\frac{3}{16}$

$P\left(\text{green from Box-I}\right)=\frac{5}{16}$

Step 3: Find P(G) — "at least one of the two chosen balls is green".

A ball is green in the pair if either b (from Box-I) is green, or the second ball drawn is green. Let's enumerate all cases where at least one ball is green:

Case A: b = red (from Box-I), second ball = green (from Box-II)

$P\left(\text{A}\right)=\frac{8}{16}\times \frac{15}{48}=\frac{1}{2}\times \frac{5}{16}=\frac{5}{32}$

Case B: b = blue (from Box-I), second ball = green (from Box-III)

$P\left(\text{B}\right)=\frac{3}{16}\times \frac{12}{16}=\frac{3}{16}\times \frac{3}{4}=\frac{9}{64}$

Case C: b = green (from Box-I), second ball = any ball from Box-IV (b itself is already green, so at least one ball is green regardless of what is drawn from Box-IV)

$P\left(\text{C}\right)=\frac{5}{16}\times 1=\frac{5}{16}$

Therefore:

$P\left(G\right)=\frac{5}{32}+\frac{9}{64}+\frac{5}{16}$

Converting to a common denominator of 64:

$P\left(G\right)=\frac{10}{64}+\frac{9}{64}+\frac{20}{64}=\frac{39}{64}$

Step 4: Find P(W ∩ G) — "at least one ball is green AND one ball is white".

White balls only exist in Box-IV. So a white ball can only be picked when b = green (from Box-I), because that triggers drawing from Box-IV. In this scenario, b itself is green, so the "at least one green" condition is automatically satisfied.

So W ∩ G happens exactly when: b = green from Box-I AND second ball = white from Box-IV.

$P(W\cap G)=\frac{5}{16}\times \frac{6}{32}=\frac{5}{16}\times \frac{3}{16}=\frac{15}{256}$

Step 5: Compute the conditional probability.

$P\left(W\right|G)=\frac{P(W\cap G)}{P\left(G\right)}=\frac{\frac{15}{256}}{\frac{39}{64}}$

Simplify by multiplying numerator by the reciprocal of the denominator:

$=\frac{15}{256}\times \frac{64}{39}=\frac{15}{4}\times \frac{1}{39}=\frac{15}{156}=\frac{5}{52}$

Therefore, the conditional probability is:

$P\left(W\right|G)=\frac{5}{52}$