A) \[\frac{(n-m)\,(n-m+1)}{(n-1)}\]
B) \[\frac{(n-m)\,(n-m+1)}{2n}\]
C) \[\frac{(n-m)\,(n-m-1)}{2n\,(n-1)}\]
D) \[\frac{(n-m)\,(n-m+1)}{2n\,(n-1)}\]
Correct Answer: D
Solution :
Let the first number be \[x\] and second is \[y.\] Let \[A\] denotes the event that the difference between the first and second number is at least \[m.\] Let \[{{E}_{x}}\] denote the event that the first number chosen is \[x,\] we must have \[x-y\ge m\] or \[y\le x-m.\] Therefore \[x>m\] and \[y<n-m.\] Thus \[P({{E}_{x}})=0\] for \[0<x\le m\] and \[P({{E}_{x}})=\frac{1}{n}\] for \[m<x\le n.\] Also \[P(A/{{E}_{x}})=\frac{(x-m)}{(n-1)}\] Therefore, \[P(A)=\sum\limits_{x=1}^{n}{P({{E}_{x}})\,\,P(A/{{E}_{x}})}\] \[=\sum\limits_{x=m+1}^{n}{P({{E}_{x}})\,\,P(A/{{E}_{x}})}=\sum\limits_{x=m+1}^{n}{\frac{1}{n}.\frac{x-m}{n-1}}\] \[=\frac{1}{n(n-1)}[1+2+3+.....+(n-m)]\] \[=\frac{(n-m)\,(n-m+1)}{2n(n-1)}.\]
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