There are three ants on a triangle, one at each corner. At a given moment in time, they all set off for a different corner at random. What is the probability that they don’t collide?
There are two methods through which we can solve this problem.
Consider the triangle ABC where we name each corner as A, B and C. There is one ant at each corner of a triangle. We can assume that the ants move towards different corners along the edges of the triangle.
Total no. of movements that the ants can make can be as follow;
A->B, B->C, C->A; A->B, B->A, C->A; A->B, B->A, C->B; A->B, B->C, C->B; A->C, B->C, C->A; A->C, B->A, C->A; A->C, B->A, C->B; A->C, B->C, C->B
Hence the total numbers of movements are 8
Non-colliding movements that the ants can make are as follows;
A->B, B->C, C->A; A->C, B->A, C->B
Hence, total Non-colliding movements are 2
(i.e. the all ants move either in the clockwise or anti-clockwise direction at the same time)
So probability of not colliding = 2/8 = 1/4
Let the three ants are a, b, c.
There are two cases when they will not collide, the one is when they all move clockwise and the other is when they all move anticlockwise.
They will collide if any two ants move towards each other, at the same time the third ant can move in clockwise or in anticlockwise. so for each pair there are 2 such cases. And there are 3 pairs possible (a,b), (b,c) and (c,a). So total 3*2 = 6 cases when they will collide.
So probability that they will not collide is 2/(2+6) i.e. 1/4
So by both the methods the probability that they don’t collide is 1/4 which is 0.25