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Question

Let the lines x+y=20 and x+y=10 meet the coordinates axes at A,B and C,D respectively. If a point M(a,b) where a,bN, is randomly selected from inside the triangle AOB, then the probability that it lies outside the triangle COD is 
(Here, O is the origin)
  1. 2738 
  2. 34
  3. 1419
  4. 1519 


Solution

The correct option is C 1419

x+y=20 
Number of points inside AOB is given by,
x+y19
17+31C31=19C2=171

Similarly, number of points inside the COD=9C2=36
So, number of points outside COD and inside AOB
=171369     [9 points on line CD]
=126

Therefore, required probability is,
P=126171=1419

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