A) \[1-\frac{2}{n}\]
B) \[\frac{2}{n-1}\]
C) \[1-\frac{1}{n}\]
D) None of these
Correct Answer: A
Solution :
Let total no. of persons = n Since, two persons are selected \[\therefore \,\,\,\,total\,\,no.of\,\,ways\,\,of\,\,selection={{ }^{n}}{{C}_{2}}\] Then \[(n -2)\] persons not selected are arranged in places shown below by circles and the selected 2 persons can be arranged at places stated by dots (Dots are \[n-1\] in number) Therefore the favourable no. of ways \[{{=}^{n-1}}{{C}_{2}}\] \[0\,\,\centerdot \,\,0\,\,\centerdot \,\,0\,\,\centerdot \,\,0\,\,\centerdot \,\,0\,\,\centerdot \,\,0\] \[\therefore \,\,\,P(not\,\,together)=\frac{^{n-1}{{C}_{2}}}{^{n}{{C}_{2}}}\] \[=\,\,\,\frac{(n-1)!}{2!(n-3)!}\times \frac{2!(n-2)!}{n!}=\frac{(n-2)(n-3)!(n-1)!}{(n-3)!\,\,\,n(n-1)!}\] \[P=\frac{(n-2)}{n}=1-\frac{2}{n}\]
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