JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Probability XII

A box

B_{1}

contains 1 white ball, 3 red balls and 2 black balls. Another box

B_{2}

contains 2 white balls, 3 red balls and 4 black balls. A third box

B_{3}

contains 3 white balls, 4 red balls and 5 black balls.

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box

B_{2}

(A) $\frac{116}{181}$
(B) $\frac{126}{181}$
(C) $\frac{65}{181}$
(D) $\frac{55}{181}$ ✓ Correct

Answer: (D)

Solution

Using Bayes' theorem:

P (B_{2} ∣ W R) = \frac{P ( W R ∣ B _{2} ) \cdot P ( B _{2} )}{P ( W R )}

where

P (B_{2}) = 1/3

(equal probability of selecting any box).

P (W R ∣ B_{1}) = \frac{( 1 1 ) \times ( 1 3 )}{( 2 6 )} = \frac{3}{15} = \frac{1}{5}

P (W R ∣ B_{2}) = \frac{( 1 2 ) \times ( 1 3 )}{( 2 9 )} = \frac{6}{36} = \frac{1}{6}

P (W R ∣ B_{3}) = \frac{( 1 3 ) \times ( 1 4 )}{( 2 12 )} = \frac{12}{66} = \frac{2}{11}

P (W R) = \frac{1}{3} (\frac{1}{5} + \frac{1}{6} + \frac{2}{11}) = \frac{1}{3} \times \frac{66 + 55 + 60}{330} = \frac{181}{990}

P (B_{2} ∣ W R) = \frac{\frac{1}{3} \times \frac{1}{6}}{\frac{181}{990}} = \frac{\frac{1}{18}}{\frac{181}{990}} = \frac{990}{18 \times 181} = \frac{55}{181}

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Probability XII

Solution

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