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 ?
See Solution : Ants on a triangle
Solution 1:
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
Solution 2 :
Consider the triangle ABC. We assume that the ants move towards different corners along the edges of the triangle.
Total no. of movements: 8
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
Non-colliding movements: 2
A->B, B->C, C->A; A->C, B->A, C->B
(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
Yes 1/4 is right… Each ant can move in 2 ways. There are 3 ants.. Hence, total number of combinations = 2x2x2 = 8. They won’t collide if they all move in same direction (all clockwise or all anticlockwise) I.e., 2 ways. So req prob = 2/8 = 1/4. 🙂
even i am getting 1/3 ..got only 6 paths…they are as follows
ab bc ca
ab ba ca
ab ba cb
ab bc cb
bc ac ca
ca ba ac
this answer is wrong… correct answer is 1/3
how??