Question
Consider a set $$P$$ containing $$n$$ elements. A subset $$A$$ of $$P$$ is drawn and there after set $$P$$ is reconstructed. Now one more subset $$B$$ of $$P$$ is drawn. Probability of drawing sets $$A$$ and $$B$$ so that $$A \cap B$$ has exactly one element is :
A.
$${\left( {\frac{3}{4}} \right)^n}.n$$
B.
$$n.{\left( {\frac{3}{4}} \right)^{n - 1}}$$
C.
$$\left( {n - 1} \right).{\left( {\frac{3}{4}} \right)^n}$$
D.
none of these
Answer :
$$n.{\left( {\frac{3}{4}} \right)^{n - 1}}$$
Solution :
Let $${x_i}$$ be any element of set $$P$$, we have following possibilities
$$\eqalign{
& \left( {\bf{i}} \right)\,\,{x_i}\, \in \,A,\,{x_i}\, \in \,B\,; \cr
& \left( {{\bf{ii}}} \right)\,\,{x_i}\, \in \,A,\,{x_i}\, \notin \,B\,; \cr
& \left( {{\bf{iii}}} \right)\,\,{x_i}\, \notin \,A,\,{x_i}\, \in \,B\,; \cr
& \left( {{\bf{iv}}} \right)\,\,{x_i}\, \notin \,A,\,{x_i}\, \notin \,B \cr} $$
Clearly, the element $${x_i}\, \in \,A \cap B$$ if it belongs to $$A$$ and $$B$$ both. Thus out of these $$4$$ ways only first way is favourable. Now the element that we want to be in the intersection can be chosen in $$'n'$$ different ways.
Hence required probability is $$n.{\left( {\frac{3}{4}} \right)^{n - 1}}.$$