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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.
One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The balls was found to be red. If the probability that this red ball was drawn from box II is 13\dfrac{1}{3} , then the correct option(s) with the possible values of n1,n2,n3andn4{n}_{1}, {n}_{2}, {n}_{3}and{n}_{4} is(are)Let n _ 1 and n _ 2 be the number of red and black balls, respectively, in box I. Let n _ 3 and n _ 4 be the number of

Options

  1. An1=3,n2=3,n3=5,n4=15{n}_{1} = 3, {n}_{2} = 3, {n}_{3} = 5, {n}_{4} = 15
  2. Bn1=3,n2=6,n3=10,n4=50{n}_{1} = 3, {n}_{2} = 6, {n}_{3} = 10, {n}_{4} = 50
  3. Cn1=8,n2=6,n3=5,n4=20{n}_{1} = 8, {n}_{2} = 6, {n}_{3} = 5, {n}_{4} = 20
  4. Dn1=6,n2=12,n3=5,n4=20{n}_{1} = 6, {n}_{2} = 12, {n}_{3} = 5, {n}_{4} = 20

Correct answer

A. n1=3,n2=3,n3=5,n4=15{n}_{1} = 3, {n}_{2} = 3, {n}_{3} = 5, {n}_{4} = 15

Step-by-step solution


Let n _ 1 and n _ 2 be the number of red and black balls, respectively, in box I. Let n _ 3 and n _ 4 be the number of
13=12(n3n3+n4)12(n1n1+n2+n3n3+n4)\dfrac{1}{3} = \dfrac{\dfrac{1}{2}\left(\dfrac{{n}_{3}}{{n}_{3} + {n}_{4}}\right)}{\dfrac{1}{2}\left(\dfrac{{n}_{1}}{{n}_{1} + {n}_{2}} + \dfrac{{n}_{3}}{{n}_{3} + {n}_{4}}\right)}
13=(n3n3+n4)(n1n1+n2+n3n3+n4)\dfrac{1}{3} = \dfrac{\left(\dfrac{{n}_{3}}{{n}_{3} + {n}_{4}}\right)}{\left(\dfrac{{n}_{1}}{{n}_{1} + {n}_{2}} + \dfrac{{n}_{3}}{{n}_{3} + {n}_{4}}\right)}
(i) RHS520520+36=1414+12=141+24=13RHS\dfrac{\dfrac{5}{20}}{\dfrac{5}{20} + \dfrac{3}{6}} = \dfrac{\dfrac{1}{4}}{\dfrac{1}{4} + \dfrac{1}{2}} = \dfrac{\dfrac{1}{4}}{\dfrac{1 + 2}{4}} = \dfrac{1}{3}
(ii) RHS10601060+39=1616+13=1612=13RHS\dfrac{\dfrac{10}{60}}{\dfrac{10}{60} + \dfrac{3}{9}} = \dfrac{\dfrac{1}{6}}{\dfrac{1}{6} + \dfrac{1}{3}} = \dfrac{\dfrac{1}{6}}{\dfrac{1}{2}} = \dfrac{1}{3}
(iii) 525814+525=1547+15=1520+735=727\dfrac{\dfrac{5}{25}}{\dfrac{8}{14} + \dfrac{5}{25}} = \dfrac{\dfrac{1}{5}}{\dfrac{4}{7} + \dfrac{1}{5}} = \dfrac{\dfrac{1}{5}}{\dfrac{20 + 7}{35}} = \dfrac{7}{27}
(iv) 525618+525=155+315=15815=15×158=38\dfrac{\dfrac{5}{25}}{\dfrac{6}{18} + \dfrac{5}{25}} = \dfrac{\dfrac{1}{5}}{\dfrac{5 + 3}{15}} = \dfrac{\dfrac{1}{5}}{\dfrac{8}{15}} = \dfrac{1}{5} \times \dfrac{15}{8} = \dfrac{3}{8} .
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