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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Probability

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Probability

Bag B_1 contains 6 white and 4 blue balls, Bag B_2 contains 4 white...

Bag B1B_1 contains 6 white and 4 blue balls, Bag B2B_2 contains 4 white and 6 blue balls, and Bag B3B_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 B2B_2, is :

Options

  1. A415\frac{4}{15}
  2. B13\frac{1}{3}
  3. C25\frac{2}{5}
  4. D23\frac{2}{3}

Correct answer

A. 415\frac{4}{15}

Step-by-step solution

E1:BagB1E_1: B a g B_1 is selected B1B2B3 6W4B 4 W6 B5 W5 B\begin{array}{lll} B_1 & B_2 & B_3 \\ \text{ 6W4B } & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B} \end{array} 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\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)} =13×41013×610+13×410+13×510=415\begin{aligned} & =\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}
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