JEE Main 2025 Jan 28 Shift 2, Mathematics Q1: Total Probability Theorem And Bayes' Theorem

JEE Main2025Jan 28, Shift 2Mathematics

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ , is :

A
$\frac{1}{3}$
B
$\frac{4}{15}$
C
$\frac{2}{3}$
D
$\frac{2}{5}$

Solution

Let $E_{i}$ be the event that bag $B_{i}$ is selected and $A$ the event that the drawn ball is white. With equal selection probabilities $P (E_{i}) = \frac{1}{3}$ , the probabilities of drawing a white ball from each bag are $\frac{6}{10}$ , $\frac{4}{10}$ , $\frac{5}{10}$ respectively.

By Bayes' theorem:

$P (\frac{E _{2}}{A}) = \frac{P ( E _{2} ) P ( A / E _{2} )}{\sum _{i} P ( E _{i} ) P ( A / E _{i} )} = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} ( \frac{6}{10} + \frac{4}{10} + \frac{5}{10} )} = \frac{4}{15}$ .

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JEE Main 2025 Jan 28 Shift 2, Mathematics Q1: Total Probability Theorem And Bayes' Theorem

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