PrepRung
Sign inSign up free

JEE Advanced 2016 Paper 2 · Mathematics · Probability · medium

A fair die is rolled. If the number obtained is 1 or 2, two balls are drawn one after another without replacement from urn U1 containing 5 red and 3…

Single correct option

A fair die is rolled. If the number obtained is 11 or 22, two balls are drawn one after another without replacement from urn U1U_1 containing 55 red and 33 blue balls; otherwise the two balls are drawn in the same way from urn U2U_2 containing 22 red and 66 blue balls. Given that the two balls drawn have the same colour, the probability that they came from U1U_1 is
  1. A. 1329\dfrac{13}{29}
  2. B. 13\dfrac{1}{3}
  3. C. 1358\dfrac{13}{58}
  4. D. 1345\dfrac{13}{45}
Show answer and solution

Answer: (D) 1345\dfrac{13}{45}

Use Bayes' theorem with P(U1)=13P(U_1)=\frac13 and P(U2)=23P(U_2)=\frac23. For U1U_1: the probability that both balls have the same colour is 5⋅4+3⋅28⋅7=2656=1328\frac{5\cdot4+3\cdot2}{8\cdot7}=\frac{26}{56}=\frac{13}{28}. For U2U_2: it is 2⋅1+6⋅58⋅7=3256=47=1628\frac{2\cdot1+6\cdot5}{8\cdot7}=\frac{32}{56}=\frac{4}{7}=\frac{16}{28}. Let EE be the event that the colours match. Then P(U1∣E)=13⋅132813⋅1328+23⋅1628=1313+32=1345P(U_1\mid E)=\dfrac{\frac13\cdot\frac{13}{28}}{\frac13\cdot\frac{13}{28}+\frac23\cdot\frac{16}{28}}=\dfrac{13}{13+32}=\dfrac{13}{45}. Option (A) forgets the prior weights 13,23\frac13,\frac23 (it equals 13/2813/28+16/28\frac{13/28}{13/28+16/28}), option (B) is the prior itself, and option (C) wrongly halves the correct numerator ratio.
Report a mistake

Original question written in the JEE Advanced 2016 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake