JEE Main 2025 Apr 2 Shift 2, Mathematics Q10: Total Probability Theorem And Bayes' Theorem

JEE Main2025Apr 2, Shift 2Mathematics

Given three indentical 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 balls 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
$7$
D
$8$

Solution

Each bag is chosen with probability $\frac{1}{3}$ .

Apply Bayes' theorem. For Red, $P (Red ∣ B_{i}) = \frac{3}{10}, \frac{4}{10}, \frac{5}{10}$ for the three bags:

$p = P (B_{I} ∣ R) = \frac{\frac{1}{3} \cdot \frac{3}{10}}{\frac{1}{3} ( \frac{3}{10} + \frac{4}{10} + \frac{5}{10} )} = \frac{3}{12} = \frac{1}{4}$ .

For Green, $P (G ∣ B_{i}) = \frac{5}{10}, \frac{3}{10}, \frac{4}{10}$ :

$q = P (B_{I I I} ∣ G) = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} ( \frac{5}{10} + \frac{3}{10} + \frac{4}{10} )} = \frac{4}{12} = \frac{1}{3}$ .

$∴ \frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$ .

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JEE Main 2025 Apr 2 Shift 2, Mathematics Q10: Total Probability Theorem And Bayes' Theorem

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