JEE Advanced 2013 · Paper 2 · Mathematics Q52

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MathematicsSingle correct+3 / -1Question 52 of 60

Passage

Paragraph for Questions 51 and 52

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 $B_2$ is

\dfrac{116}{181}

\dfrac{126}{181}

\dfrac{65}{181}

\dfrac{55}{181}

Answer and worked solution

Official answer: D

One white and one red from each box:

$B_1$ : $\dfrac{1\cdot3}{\binom62}=\dfrac15$
$B_2$ : $\dfrac{2\cdot3}{\binom92}=\dfrac16$
$B_3$ : $\dfrac{3\cdot4}{\binom{12}2}=\dfrac2{11}$

Bayes (boxes equally likely):
$P(B_2\mid WR)=\frac{\frac16}{\frac15+\frac16+\frac2{11}}=\frac{55}{66+55+60}=\frac{55}{181}.$

(D)

Topics:Bayes' theorem

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JEE Advanced 2013 · Paper 2, Mathematics question 52