Verified Solution • 2025–26 Session moderate

QJEEM25Maae95: JEE Main 2025 Mathematics - Class 12 JEEM25Maae95 Solution: Bag contains 6 white and 4 blue balls, Bag contains 4 white and …

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:

Multiple Choice Options

  • A 25\frac{2}{5}
  • B 415\frac{4}{15}
  • C 13\frac{1}{3}
  • D 23\frac{2}{3}

Step-by-Step Verified Solution

E1:BagB1E_1: B a g B_1 is selected

B1amp;B2amp;B36 W4 Bamp;4 W6 Bamp;5 W5 B\begin{array}{lll}B_1 & B_2 & B_3 \\6 \mathrm{~W} 4 \mathrm{~B} & 4 \mathrm{~W} 6 \mathrm{~B} & 5 \mathrm{~W} 5 \mathrm{~B}\end{array}

E2:E_2: bag B2B_2 is selected

E3:BagB3E_3: B a g B_3 is selected

A : Drawn ball is white

We have to find P(E2A)P\left(\frac{E_2}{A}\right)

amp;P(E2A)=P(E2)P(AE2)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)amp;=13×41013×610+13×410+13×510amp;=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}

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