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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Mathematics
Chapter
Probability

JEE Advanced 2015 · Paper 2 · Mathematics · Probability

Let n _ 1 and n _ 2 be the number of red and black balls,...

Let n1andn2{n}_{1}and{n}_{2} be the number of red and black balls, respectively, in box I. Let n3andn4{n}_{3}and{n}_{4} be the number of red and black balls, respectively, in box II.
A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 13\dfrac{1}{3} , then the correct options(s) with the possible values of n1andn2{n}_{1}and{n}_{2} is(are)

Options

  1. An1=4andn2=6{n}_{1} = 4and{n}_{2} = 6
  2. Bn1=2andn2=3{n}_{1} = 2and{n}_{2} = 3
  3. Cn1=10andn2=20{n}_{1} = 10and{n}_{2} = 20
  4. Dn1=3andn2=6{n}_{1} = 3and{n}_{2} = 6

Correct answer

C. n1=10andn2=20{n}_{1} = 10and{n}_{2} = 20

Step-by-step solution

Probability = \dfrac{{n}_{1}}{{n}_{1} + {n}_{2}} ⋅ \dfrac{{n}_{1} - 1}{{n}_{1} + {n}_{2} - 1} + \dfrac{{n}_{2}}{{n}_{1} + {n}_{2}} ⋅ \dfrac{{n}_{1}}{{n}_{1} + {n}_{2} - 1} = \dfrac{1}{3}
⇒n1(n1−1)+n2n1(n1+n2)(n1+n2−1)=13⇒ \dfrac{{n}_{1}\left({n}_{1} - 1\right) + {n}_{2}{n}_{1}}{\left({n}_{1} + {n}_{2}\right) ( {n}_{1} + {n}_{2} - 1 )} = \dfrac{1}{3}
⇒n1=10,n2=20⇒ {n}_{1} = 10, {n}_{2} = 20
or n1=3,n2=6{n}_{1} = 3, {n}_{2} = 6
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