JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 14)
Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $${1 \over {12}},$$ then the number of children in each family is :
3
4
5
6
Explanation
Let the number of children in each family be x.
Thus the total number of children in both the families are 2x
Now, it is given that 3 tickets are distributed amongst the children of these two families.
Thus, the probability that all the three tickets go to the children in family B
= $${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$$ = $${1 \over {12}}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$$ = $${1 \over {12}}$$
$$ \Rightarrow $$ $${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$$ = $${1 \over 6}$$
Thus, the number of children in each family is 5.
Thus the total number of children in both the families are 2x
Now, it is given that 3 tickets are distributed amongst the children of these two families.
Thus, the probability that all the three tickets go to the children in family B
= $${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$$ = $${1 \over {12}}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$$ = $${1 \over {12}}$$
$$ \Rightarrow $$ $${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$$ = $${1 \over 6}$$
Thus, the number of children in each family is 5.
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