CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let $$A$$ be a set containing $$n$$ elements. A subset $$P$$ of the set $$A$$ is chosen at random.
The set $$A$$ is reconstructed by replacing the element of $$P$$, and another subsets $$Q$$ of $$A$$ is chosen at random. The probability that $$\left( P\cap Q \right)  $$ contains exactly $$m(m<n)$$ elements is


A
3nm4n
loader
B
nCm3m4n
loader
C
nCm3nm4n
loader
D
none of these
loader

Solution

The correct option is C $$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } } $$
We know that the number of subsets of a set containing $$n$$ elements is $${ 2 }^{ n }$$.
Therefore, the number of ways of choosing $$P$$ and $$Q$$ is $$^{ { 2 }^{ n } }{ { C }_{ 1 } }\times _{  }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$$.
Out of $$n$$ elements, $$m$$ elements can be chosen in $$^{ n }{ { C }_{ m } }$$ ways.
If $$\left( P\cap Q \right) $$ contains exactly in elements, 
then from the remaining $$n-m$$ elements either an element belongs to $$P$$ or $$Q$$ but not both $$P$$ and $$Q$$.
Suppose $$P$$ contains $$r$$ elements from the remaining $$n-m$$ elements.
Then $$Q$$ may contain any number of elements from the remaining $$(n-m)-r$$ elements. 
Therefore, $$P$$ and $$Q$$ can be chosen in $$^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r }$$
But $$r$$ can vary from $$0$$ to $$(n-m)$$. So, $$P$$ and $$Q$$ can be chosen in general in
$$\displaystyle \left( \sum _{ r=0 }^{ n-m }{ ^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r } }  \right) _{  }^{ n }{ { C }_{ m } }={ \left( 1+2 \right)  }^{ n-m }\times _{  }^{ n }{ { C }_{ m } }=_{  }^{ n }{ { C }_{ m } }\times { 3 }^{ n-m }$$
Hence, required probability $$\displaystyle =\dfrac { ^{ n }{ { C }_{ m } }\times { 3 }^{ n-m } }{ { 4 }^{ n } } $$

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image