A) ½
B) \[1/({{2}^{k}}-1)\]
C) \[1/{{2}^{k}}\]
D) \[1/{{3}^{k}}\]
Correct Answer: B
Solution :
| \[n\left( x \right)=k,\] number of subsets of \[X={{2}^{k}}\] |
| The number of elements and the binomial expansion \[{{2}^{k}}=\,\,{{\,}^{k}}\,\,{{C}_{0}}+{}^{k}{{C}_{1}}+{}^{k}{{C}_{2}}+...+{}^{k}{{C}_{k-2}}+{}^{k}{{C}_{k-1}}+{}^{k}{{C}_{k}}\] |
| \[^{k}{{C}_{0}}=\] null set (without any element) and \[^{k}{{C}_{k}}\](universal set) are complementary. Similarly there are \[k\] singletons \[{{(}^{k}}{{C}_{1}})\] which will have \[k\]set with \[(k-1)\]elements each as their complementary sets. |
| \[\therefore \]No. of such combination \[=1/2({{2}^{k}})\] |
| Now two subsets from \[{{2}^{k}}\] subsets can be selected in \[{}^{{{2}^{k}}}{{C}_{2}}\]ways |
| \[\therefore \]Required probability\[P(E)=\frac{{{2}^{k}}}{2{{\cdot }^{{{2}^{k}}}}{{C}_{2}}}\] |
| Put \[{{2}^{k}}=m\] |
| So, \[P(E)=\frac{m\cdot 2}{2\cdot m(m-1)}=\frac{1}{m-1}=\frac{1}{({{2}^{k}}-1)}\] |
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