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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

JEE Advanced 2013 · Paper 2 · Mathematics · Probability

Paragraph: A box B_ 1 contains 1 white ball, 3 red balls and 2...

Paragraph: A box B1B_{1} contains 11 white ball, 33 red balls and 22 black balls. Another box B2B_{2} contains 22 white balls, 33 red balls and 44 black balls. A third box B3B_{3} contains 33 white balls, 44 red balls and 55 black balls.
Question: If 22 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 22 balls are drawn from box B2B_{2} is

Options

  1. A116181\dfrac{1 1 6}{1 8 1}
  2. B126181\dfrac{1 2 6}{1 8 1}
  3. C65181\dfrac{6 5}{1 8 1}
  4. D55181\dfrac{5 5}{1 8 1}

Correct answer

D. 55181\dfrac{5 5}{1 8 1}

Step-by-step solution

Let A:A: one ball is white and other is red
E1:{E}_{1}: both balls are from box B1{B}_{1}
E2:{E}_{2}: both ball are from box B2{B}_{2}
E3:{E}_{3}: both ball are from box B3{B}_{3}
Hence, P(required)=P(E2A)\text{P}\left(\text{required}\right) = \text{P}\left(\dfrac{{\text{E}}_{2}}{\text{A}}\right)
=P(AE2)⋅P(E2)P(AE1)⋅P(E1)+P(AE2)⋅P(E2)+P(AE3)⋅P(E3)= \dfrac{\text{P} \left(\dfrac{\text{A}}{{\text{E}}_{2}}\right) \cdot \text{P} \left({\text{E}}_{2}\right)}{\text{P} \left(\dfrac{\text{A}}{{\text{E}}_{1}}\right) \cdot \text{P} \left({\text{E}}_{1}\right) + \text{P} \left(\dfrac{\text{A}}{{\text{E}}_{2}}\right) \cdot \text{P} \left({\text{E}}_{2}\right) + \text{P} \left(\dfrac{\text{A}}{{\text{E}}_{3}}\right) \cdot \text{P} \left({\text{E}}_{3}\right)}
=2C1×3C19C2×131C1×3C16C2×13+2C1×3C19C2×13+3C1×4C112C2×13= \dfrac{\dfrac{{}^{2} {\text{C}}_{1} \times {}^{3} {\text{C}}_{1}}{{}^{9} {\text{C}}_{2}} \times \dfrac{1}{3}}{\dfrac{{}^{1} {\text{C}}_{1} \times {}^{3} {\text{C}}_{1}}{{}^{6} {\text{C}}_{2}} \times \dfrac{1}{3} + \dfrac{{}^{2} {\text{C}}_{1} \times {}^{3} {\text{C}}_{1}}{{}^{9} {\text{C}}_{2}} \times \dfrac{1}{3} + \dfrac{{}^{3} {\text{C}}_{1} \times {}^{4} {\text{C}}_{1}}{{}^{1 2} {\text{C}}_{2}} \times \dfrac{1}{3}}
=1615+16+211= \dfrac{\dfrac{1}{6}}{\dfrac{1}{5} + \dfrac{1}{6} + \dfrac{2}{1 1}}
=55181= \dfrac{5 5}{1 8 1}
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