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,b∈N, is randomly selected from inside the triangle AOB, then the probability that it lies outside the triangle COD is

(Here, O is the origin)

(Here, O is the origin)

- 2738
- 34
- 1419
- 1519

Solution

The correct option is **C** 1419

x+y=20

Number of points inside △AOB is given by,

x+y≤19

17+3−1C3−1=19C2=171

Similarly, number of points inside the △COD=9C2=36

So, number of points outside △COD and inside △AOB

=171−36−9 [9 points on line CD]

=126

Therefore, required probability is,

P=126171=1419

x+y=20

Number of points inside △AOB is given by,

x+y≤19

17+3−1C3−1=19C2=171

Similarly, number of points inside the △COD=9C2=36

So, number of points outside △COD and inside △AOB

=171−36−9 [9 points on line CD]

=126

Therefore, required probability is,

P=126171=1419

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