JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Probability XII

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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

(A)(A) $\frac{15}{256}$
(B)(B) $\frac{3}{16}$
(C)(C) $\frac{5}{52}$ ✓ Correct
(D)(D) $\frac{1}{8}$

Answer: (C)

Solution

Let A = "one of the chosen balls is white" and B = "at least one of the chosen balls is green".

We need

P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}

.

**Event

A \cap B

**: One ball is white and at least one is green. This happens only when:

First ball from Box-I is green (probability $\frac{5}{16}$ )
Second ball from Box-IV is white (probability $\frac{6}{32}$ )

P (A \cap B) = \frac{5}{16} \times \frac{6}{32} = \frac{30}{512} = \frac{15}{256}

Event B: At least one ball is green. This happens when:

First ball is green (from Box-I): $\frac{5}{16} \times 1 = \frac{5}{16}$
First ball is red, second is green (from Box-II): $\frac{8}{16} \times \frac{15}{48} = \frac{8}{16} \times \frac{5}{16} = \frac{5}{32}$
First ball is blue, second is green (from Box-III): $\frac{3}{16} \times \frac{12}{16} = \frac{36}{256} = \frac{9}{64}$

P (B) = \frac{5}{16} + \frac{5}{32} + \frac{9}{64} = \frac{20 + 10 + 9}{64} = \frac{39}{64}

Therefore:

P (A ∣ B) = \frac{15/256}{39/64} = \frac{15}{256} \times \frac{64}{39} = \frac{15}{4 \times 39} = \frac{15}{156} = \frac{5}{52}

The answer is (C).

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Single correct option

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Probability XII

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