Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.
Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.
Number of points
Plane
Probability
But try the problem first...
Answer: is 489.
AIME I, 1999, Question 10
Geometry Vol I to IV by Hall and Stevens
First hint
\(10 \choose 3\) sets of 3 points which form triangles,
Second Hint
fourth distinct segment excluding 3 segments of triangles=45-3=42
Final Step
Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)
where \({45 \choose 4}\) is choosing 4 segments from 45 segments
=\(\frac{16}{473}\) then m+n=16+473=489.
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