Two objects are placed on opposite corners of a cube. The

You are right. I forgot to multiply by 6 in my approach. Answer is 1/6.
+1

we have 6 points.
the total number of different combinations is 6(first object)*6(second object)=36
the number of combinations with collision is 6 (not 1 as I thought)
Therefore,
\(p = \frac{6}{36} = \frac{1}{6}\)

You need minimum of 3 steps to get to the opposite corners. 1 step along x,y,and z axis each. The total number of routes is
3!/(1!*1!*1!) = 6.

The other guy has 6 routes too.
If others' route is fixed, then 1/6 of your routes will lead to collision.

so p=1/6

Two objects are placed on opposite corners of a cube. The objects are designed only to move along the edges of the cube and are equipped with a power device which allows them to do so at the same constant speed of S inches per minute. If both objects begin moving at the same time, each taking one of the shortest paths towards the initial position of the other, what is the probability that the two objects will collide?

Ans: Each object has 6 ways to reach the other object's initial position. (passing through 3 connected edges)
Hence total number of possibilities = 6 * 6 = 36

Of these 36 ways , collision will occur if each object chooses the same path, i.e both should follow one of the possible 6 ways
= 6
Probability that they will collide = 6/36 = 1/6
=

We need the possible short paths from the opposite corners.

The object would need to travel 2 times horizontal and once vertical to reach the opposite corner via the shortest path. Hence total number of paths = 3! (arrange HHV in diff comb).

The probability of collision can be if both the objects take the same path. Therefore = 1 / 3! = 1/6

Two objects are placed on opposite corners of a cube. The objects are designed only to move along the edges of the cube and are equipped with a power device which allows them to do so at the same constant speed of S inches per minute. If both objects begin moving at the same time, each taking one of the shortest paths towards the initial position of the other, what is the probability that the two objects will collide?



Tell me if this is the right method too. This method that I took seemed simpler when compared.

There are 6 shortest ways to travel to the opposite corner of the cube.
First object can take route in 6/6 ways.

If the second object has to collide into this, it has to choose the one route that the first object has chosen. So in 1/6 ways.

Therefore total probability = (6/6)*(1/6) = 1/6


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I used the following approach, though don't no is it right.
Any object has to go 3 points to reach the opposite corner.
The total number of ways is 3*2*1=6.
Hence, two objects will collide if they chose the same route and the chance is 1/6.

good question ... the ans is 1/6..

the number of possible ways to reach to other end for object A = 6
the number of possible ways to reach to other end for object B = 6

so the probability of meeting each other = (1/6)(1/6)=1/36

the number of possible start point =6

so the probability is 6/36 = 1/6

My ans is 1/6..bt i solved in another way...
there are 6 ways by which any object move from one end to its opp end(shortest way)..
and if both object has to collide thn they will move on same way..means there is only one way..
so prob is 1/6

My Solution:

So object 1 has 3 steps to go to object 2's initial position. Object one can choose from among 3 edges, from that from two edges then with 1 edge so 3 x 2 x 1 = 6. Object 2 has the same 6 possible routes.

6 x 6 = 36 all possible comb. routes of object 1 and 2

For them to collide, 6 ways for them to collide.

6/36 = 1/6

I find this one tough

There are six possible paths:

The first object's path could be picked in 6 ways: 6/6
The second object's path needs to pick the same path: 1/6
Probability that they're both on the same path: 6/6 * 1/6
= 1/6 = 16.67%

Say, from opposite corner A, we can reach diagonally opposite corner B, with three hops -> 1 along (x direction), 1 along (y direction), 1 along (z direction) => total ways to reach = permutation of (x,y,z) = 3! = 6

Likewise, from opposite corner B, we can reach diagonally opposite corner A, with three hop => total ways = 3! = 6

So total ways for each point to reach each other's initial point = 6 * 6 = 36

out of this, if out of 6 ways, say A, takes x->z->y and then point B, must take y->z->x to collide.

So for each one of 6 ways, point A will take, point B must take one of its 6 ways to collide => total favourable = 6

Required probability = 6/36 = 1/6

Dear Moderators,

Could you please add the 5 answer choices so that we can solve this as a normal PS question and not just a practice question of probability chapter?

chetan2u, Gladiator59, VeritasKarishma, Bunuel, generis

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Done.

The probability of choosing a shortest path by either object= 1/3*1/2=1/6

The probability of choosing the same path by both objects= 1/6*1/6=1/36

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