JEE Advanced 2016 Paper 1 · Mathematics · Probability · medium

A fair die is thrown once. If the number is 1 or 2, two balls are drawn one after another without replacement from bag A containing 3 red and 2 blue…

Single correct option

A fair die is thrown once. If the number is

1

2

, two balls are drawn one after another without replacement from bag

A

containing

3

red and

2

blue balls. Otherwise, two balls are drawn one after another with replacement from bag

B

containing

2

red and

4

blue balls. Given that both balls drawn are red, the probability that bag

A

was used is

Show answer and solution

Answer: (C) $\dfrac{27}{47}$

Use Bayes' theorem. Bag

A

is chosen with probability

1/3

and bag

B

with probability

2/3

. From

A

without replacement,

P(RR)=\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}

. From

B

with replacement,

P(RR)=\left(\frac{2}{6}\right)^2=\frac19

. The total probability is

P(RR)=\frac13\cdot\frac{3}{10}+\frac23\cdot\frac19=\frac{1}{10}+\frac{2}{27}=\frac{47}{270}

. Therefore

P(A\mid RR)=\dfrac{1/10}{47/270}=\dfrac{27}{47}

. The value

1/3

is the prior and ignores the evidence,

9/47

is a distractor that does not follow from the data, and

20/47

is the posterior probability for bag

B

, i.e.

1-\frac{27}{47}

Original question written in the JEE Advanced 2016 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