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JEE Advanced 2024 Paper 2 · Mathematics · Probability · medium

A fair die is thrown once. If the result is 1 or 2, two balls are drawn without replacement from Urn I containing 3 red and 2 blue balls; otherwise…

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A fair die is thrown once. If the result is 11 or 22, two balls are drawn without replacement from Urn I containing 33 red and 22 blue balls; otherwise two balls are drawn without replacement from Urn II containing 22 red and 44 blue balls. Given that the two balls drawn are of the same colour, the probability that Urn I was used is
  1. A. 13\dfrac{1}{3}
  2. B. 310\dfrac{3}{10}
  3. C. 215\dfrac{2}{15}
  4. D. 720\dfrac{7}{20}
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Answer: (B) 310\dfrac{3}{10}

P(same colour∣I)=(32)+(22)(52)=410=25P(\text{same colour}\mid I)=\dfrac{\binom32+\binom22}{\binom52}=\dfrac{4}{10}=\dfrac25 and P(same∣II)=(22)+(42)(62)=715P(\text{same}\mid II)=\dfrac{\binom22+\binom42}{\binom62}=\dfrac{7}{15}. With P(I)=13P(I)=\tfrac13, P(II)=23P(II)=\tfrac23: P(I∩same)=215P(I\cap\text{same})=\tfrac{2}{15} and P(II∩same)=1445P(II\cap\text{same})=\tfrac{14}{45}. By Bayes' theorem P(I∣same)=6/456/45+14/45=310P(I\mid\text{same})=\dfrac{6/45}{6/45+14/45}=\dfrac{3}{10}. The value 215\frac{2}{15} is the joint probability (no normalisation) and 13\frac13 is the prior.
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