QJEEAdv15Ma5a19: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma5a19 Solution: Let and be the number of red and black balls, respectively, in…

Box I : red balls $\to$ n₁

black balls $\to$ n₂

Box II : red balls $\to$ n₃

black balls $\to$ n₄

$P(R) = {1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}$

P(Box II / R) $= {{{1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}} \over {{1 \over 2}.{{{n_1}} \over {{n_1} + {n_2}}} + {1 \over 2}.{{{n_3}} \over {{n_3} + {n_4}}}}}$

$= {1 \over {1 + \left( {{{{{{n_1}} \over {{n_1} + {n_2}}}} \over {{{{n_3}} \over {{n_3} + {n_4}}}}}} \right)}} = {1 \over 3}$

Thus, ${{{n_1}} \over {{n_1} + {n_2}}} = 2{{{n_3}} \over {{n_3} + {n_4}}}$

i.e. $2\left( {1 + {{{n_2}} \over {{n_1}}}} \right) = 1 + {{{n_4}} \over {{n_3}}}$ i.e. ${{{n_4}} \over {{n_3}}} - 2{{{n_2}} \over {{n_1}}} = 1$