A) \[\frac{^{2n}{{C}_{n}}}{{{2}^{2n}}}\]
B) \[\frac{1}{^{2n}{{C}_{n}}}\]
C) \[\frac{1\,.\,3\,.\,5......(2n-1)}{{{2}^{n}}}\]
D) \[\frac{{{3}^{n}}}{{{4}^{n}}}\]
Correct Answer: A
Solution :
We know that the number of sub-sets of a set containing \[n\] elements is \[{{2}^{n}}.\] Therefore the number of ways of choosing \[A\] and \[B\] is \[{{2}^{n}}.\] \[{{2}^{n}}={{2}^{2n}}\] We also know that the number of sub-sets (of X) which contain exactly \[r\] elements is \[{}^{n}{{C}_{r}}.\] Therefore the number of ways of choosing \[A\] and \[B,\] so that they have the same number elements is \[{{({}^{n}{{C}_{0}})}^{2}}+{{({}^{n}{{C}_{1}})}^{2}}+{{({}^{n}{{C}_{2}})}^{2}}+\,......+{{({}^{n}{{C}_{n}})}^{2}}={}^{2n}{{C}_{n}}\] Thus the required probability \[=\frac{{}^{2n}{{C}_{n}}}{{{2}^{2n}}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
