JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Probability

Given three identical bags each containing 10 balls, whose colours are as follows:

	Red	Blue	Green
Bag I	3	2	5
Bag II	4	3	3
Bag III	5	1	4

A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is p and if the ball is Green, the probability that it is from bag III is q, then the value of

(\frac{1}{p} + \frac{1}{q})

is:

(A)6
(B)9
(C)7✓ Correct
(D)8

Answer: (C)

Solution

Each bag is equally likely, with prior

\frac{1}{3}

. By Bayes' theorem,

p = P (Bag I ∣ Red) = \frac{\frac{1}{3} \cdot \frac{3}{10}}{\frac{1}{3} ( \frac{3}{10} + \frac{4}{10} + \frac{5}{10} )} = \frac{3}{12} = \frac{1}{4}

, so

\frac{1}{p} = 4

. Likewise

q = P (Bag III ∣ Green) = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} ( \frac{5}{10} + \frac{3}{10} + \frac{4}{10} )} = \frac{4}{12} = \frac{1}{3}

, so

\frac{1}{q} = 3

. Therefore

\frac{1}{p} + \frac{1}{q} = 7

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Probability

Solution

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