  Question

A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstrucetd by replacing the element of P. A subset Q of A is again chosen at random. Find the probability that P & Q have no common elements.

A
(3/4)n  B
(3/8)n  C
(3/4)n1  D
(3/4)n+1  Solution

The correct option is B (3/4)$$^n$$Sample space = number of ways in which we can form set A and number of ways in which we can form set B it is $${ 2 }^{ n }$$ in both the cases $$\left( ^{ n }{ { C }_{ 0 }+ }{ ^{ n }{ { C }_{ 1 }+ } }^{ n }{ { C }_{ 2 }+...^{ n }{ { C }_{ n } } } \right)$$ $$^{ n }{ { C }_{ 0 } }$$ when the subset is null set$$^{ n }{ { C }_{ 1 } }$$ when the subset contain 1 element it goes on when the subset contain all the element of the representedSo sample space $$={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$$Now number of favorable way when P subset contain no element and Q subset contain n-1 elementsP subsets contain r elements and Q subsets contain n-r elements ....p subset contain n element and q subset cantain no elements $$\sum _{ r=0 }^{ n }{ ^{ n }{ { C }_{ r } } } \left( { 2 }^{ n-r } \right) ={ 3 }^{ n }$$So the probability is $$\displaystyle { \left( \frac { 3 }{ 4 } \right) }^{ n }$$Maths

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