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Mathematics · Probability

JEE Main 2025 — 28 January, Evening Shift — Question 1

Bag B1B_{1} contains 6 white and 4 blue balls, Bag B2B_{2} contains 4 white and 6 blue balls, and BagB3\mathrm{Bag} \mathrm{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⁡B2\operatorname{Bag} \mathrm{B}_{2}, is :

  1. Option A:

    13\frac{1}{3}

  2. Option B:

    415\frac{4}{15}

    Correct
  3. Option C:

    23\frac{2}{3}

  4. Option D:

    25\frac{2}{5}

Answer: B

Step-by-step solution

E1:Bag⁡B1\mathrm{E}_{1}: \operatorname{Bag} \mathrm{B}_{1} is selected

figure

E2:\mathrm{E}_{2}: Bag B2\mathrm{B}_{2} is selected

E3:BagB3E_{3}: B a g B_{3} is selected

A : Drawn ball is white

We have to find P(E2 A)\mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{~A}}\right)

P(E2A)=P(E2)P(AE2)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)=13×41013×610+13×410+13×510=415\begin{aligned} & P\left(\frac{E_{2}}{A}\right)=\frac{P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right)}{P\left(E_{1}\right) P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) P\left(\frac{A}{E_{2}}\right)+P\left(E_{3}\right) P\left(\frac{A}{E_{3}}\right)} \\& =\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{6}{10}+\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}} \\& =\frac{4}{15} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Evening Shift
Subject
Mathematics
Chapter
Probability
Topic
Total Probability and Baye's Theorem

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