Verified Solution • 2025–26 Session moderate

QJEEM22Ma6fe5: JEE Main 2022 Mathematics - Class 12 JEEM22Ma6fe5 Solution: Bag I contains 3 red, 4 black and 3 white balls and Bag II…

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :

Multiple Choice Options

  • A 49
  • B 518
  • C 16
  • D 310

Step-by-Step Verified Solution

Let E→E \rightarrow Ball drawn from Bag II is black.

ER→E_{R} \rightarrow Bag I to Bag II red ball transferred.

EB→E_{B} \rightarrow Bag I to Bag II black ball transferred.

Ew→E_{w} \rightarrow Bag I to Bag II white ball transferred.

P(ER/E)=P(E/ER)⋅P(ER)P(E/ER)P(ER)+P(E/EB)P(EB)+P(E/EW)P(EW)P\left(E_{R} / E\right)=\frac{P\left(E / E_{R}\right) \cdot P\left(E_{R}\right)}{P\left(E / E_{R}\right) P\left(E_{R}\right)+P\left(E / E_{B}\right) P\left(E_{B}\right)+P\left(E / E_{W}\right) P\left(E_{W}\right)}

Here,

P(ER)=3/10,P(EB)=4/10,P(EW)=3/10P\left(E_{R}\right)=3 / 10, \quad P\left(E_{B}\right)=4 / 10, \quad P\left(E_{W}\right)=3 / 10

and

P(E/ER)=5/10,P(E/EB)=6/10,P(E/EW)=5/10∴P(ER/E)=15/10015/100+24/100+15/100=1554=518

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