PrepRung
Sign inSign up free

JEE Advanced 2020 Paper 1 · Mathematics · Probability · medium

Box I contains 3 red and 2 blue balls; Box II contains 4 red and 1 blue ball. One ball is chosen at random from Box I and placed in Box II. Then two…

Single correct option

Box I contains 3 red and 2 blue balls; Box II contains 4 red and 1 blue ball. One ball is chosen at random from Box I and placed in Box II. Then two balls are drawn at random without replacement from Box II and both turn out to be red. The probability that the ball transferred from Box I was red is
  1. A. 35\dfrac{3}{5}
  2. B. 23\dfrac{2}{3}
  3. C. 57\dfrac{5}{7}
  4. D. 1013\dfrac{10}{13}
Show answer and solution

Answer: (C) 57\dfrac{5}{7}

Let RR be the event that a red ball is transferred and EE the event that both balls drawn from Box II are red. P(R)=3/5P(R)=3/5 and P(Rˉ)=2/5P(\bar R)=2/5. After a red transfer Box II has 5 red and 1 blue ball, so P(E∣R)=(52)/(62)=10/15=2/3P(E\mid R)=\binom52/\binom62=10/15=2/3. After a blue transfer it has 4 red and 2 blue, so P(E∣Rˉ)=(42)/(62)=6/15=2/5P(E\mid\bar R)=\binom42/\binom62=6/15=2/5. By Bayes' theorem, P(R∣E)=(3/5)(2/3)(3/5)(2/3)+(2/5)(2/5)=10/2510/25+4/25=57P(R\mid E)=\dfrac{(3/5)(2/3)}{(3/5)(2/3)+(2/5)(2/5)}=\dfrac{10/25}{10/25+4/25}=\dfrac{5}{7}. The prior 3/53/5 is the tempting wrong answer, and 2/32/3 ignores the competing blue-transfer case.
Report a mistake

Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake