Question

Probability. Form a Triangle. .

ABCD is a square. Three points P, Q and R are chosen randomly from the four edges.

How many are the permutations of the edges selected ? [4,8,12,16,20]

How many top permutations result in PQR being a triangle? [4,8,12,16,20]

What's the probability that PQR is a triangle? [0.5, 0.75, 0.333,0.2, 1.0]

Answer 1

3 points P Q & R can occur all on one side, two one one side and one on a different side, and finally all points on different sides.

All points on one side can occur in 4 ways. No triangle is possible then.

2 points on one side, 1 on another, 0 on remaining sides. Num of Permutations is: 4!/2! = 12 .

Each of 3 points on distinct sides. There are 4 permutations.

So probability = (12×4)/(12+4+4) =3/4