JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 6)
One of the two boxes, box $${\rm I}$$ and box $${\rm I}{\rm I},$$ was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $${\rm I}{\rm I}$$ is $${1 \over 3},$$ then the correct option(s) with the possible values of $${n_1}$$ $${n_2},$$ $${n_3}$$ and $${n_4}$$ is (are)
$${n_1} = 3,{n_2} = 3,{n_3} = 5,{n_4} = 15$$
$${n_1} = 3,{n_2} = 6,{n_3} = 10,{n_4} = 50$$
$${n_1} = 8,{n_2} = 6,{n_3} = 5,{n_4} = 20$$
$${n_1} = 6,{n_2} = 12,{n_3} = 5,{n_4} = 20$$
Explanation
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. $${{{n_4}} \over {{n_3}}} - 2{{{n_2}} \over {{n_1}}} = 1$$
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