Math Is Fun Forum

kyr_leo

Member

Registered: 2016-02-23

Posts: 6

4 points

Find the probability for the following 4 points (0, y1, 0) (60, y2 ,0) (60, y3, 40) (0, y4, 40) to be co-planar, if y1, y2, y3 and y4 can take any integer value from 91 to 98.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 4 points

Hi

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

kyr_leo

Member

Registered: 2016-02-23

Posts: 6

Re: 4 points

How?

Hi

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 4 points

How?

Before I go into a fairly long and complicated EM method, I would ask if the answer given is correct.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Grantingriver

Member

Registered: 2016-02-01

Posts: 129

Re: 4 points

If we draw these points on a cartesian coordinate system, we will have a cuboid that intersects the zx plane in a rectangle. Now imagine a plane which intersects this cuboid above the zx plane and consider the four points which constitute the corners of the cross section. The second coordinates of these points has a special property: if we divide them into pairs, we could form the two pairs

and

such that each pair has the same difference to the other, so this is an essensial property for the all points to lie in the same plane. This claim can be proven by noticing that the four points, in three dimensions space, form a tetrahedron and in the spetial case where the four points lie on the same plain, the volume of the tetrahedron will be "Zero" hence:

.

Now if we consider the possibilities of the valses of the second coordinates (taking into account the premited values in the integral interval [91;98]), we will have:

1) all the four are equals.
2) three of them are equal and the third is distinct.
3) two are eqaul and two are distincts, but they have the same value.
4) two are eqaul and also two are different, but they have distinct values.
5) all of them are distinct.

In the first case. It is obvious that the differences of any pair are equal. In the second possibility we cannot form the two pairs

and

so that one of them has difference equal to the other hence they cann't lie on the same plane. In the third possible case, it is clear that we could form such pairs while in the fourth possibility it is impossible. Finally in the fifth case, we can form such pairs in many possible cases. Therefore if we denote the favorable cases "fc" we have in each case the following:

So since the probability is defined as the ratio of the favorable cases (fc) to the possible caces (pc), in other words:

Then

Note: we compare the differeces of the two pairs instead of their ratios because in the third case all of the four points will lie on the same plane if and olny if the two pairs

and

of their second components have eqaual differences (the proportionality holds in all the possible cases except the last one while the differences criterion applicable in all of them). Finally, in the first four cases the permutations can be computed easly. However, in the fifth case (where the situation is more complicated), you can use a programming language to compute the permutations. If you use "Python", the following code will handle the problem:

import itertools
list2Perm = list(itertools.permutations([91,92,93,94,95,96,97,98], 4)]
number=0
listPerm=[]
While number != len(list2Perm):
        list3Perm = list(itertools.chain(list2Perm[number]))
number += 1
        if list3Perm[0]-list3perm[1]==list3Perm[3]-list3Perm[2]:
listPerm.append(list3Perm)

print(len(listPerm))

input("\n\npress enter key to exit.")

Q.E.F

kyr_leo

Member

Registered: 2016-02-23

Posts: 6

Re: 4 points

Hi bobby; yes, your answer is correct. I would appreciate if you could post your solution.

How?

Before I go into a fairly long and complicated EM method, I would ask if the answer given is correct.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 4 points

Hi;

There are other ways to attack this problem but from a computational standpoint this is my favorite.

Four points are coplanar iff the determinant of the following matrix is 0.

http://mathworld.wolfram.com/Coplanar.html

Expanding the determinant and setting it to 0 we get:

Dividing by 2400

This is a linear Diophantine equation. The number of solutions to them can be solved by the use of generating functions.

The number of solutions of

with the constraints 

is equal to the coefficient of x^0 ( the constant term ) in the expansion of

This product is nothing more than algebra and can be done by hand methods but it is best done by computer. If you need to I will show you how to take it over to Alpha. This is the power of generating functions, a hard problem has been reduced to a computational one. The expansion of the above product yields:

We see that the constant term is 344 and so the answer is  344 / 8^4 = 43 / 512.

Now I am not saying this is the simplest solution to this problem but these methods have the virtue of being able to solve a great many combinatorics problems with little or no change.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

kyr_leo

Member

Registered: 2016-02-23

Posts: 6

Re: 4 points

Well, I do not understand the second part with the fractions, but I did it with Excel (generating all possible combinations with repetition) and got the same result

MANY THANKS!!

Hi;

There are other ways to attack this problem but from a computational standpoint this is my favorite.

Four points are coplanar iff the determinant of the following matrix is 0.

http://mathworld.wolfram.com/Coplanar.html

Expanding the determinant and setting it to 0 we get:

Dividing by 2400

This is a linear Diophantine equation. The number of solutions to them can be solved by the use of generating functions.

The number of solutions of

with the constraints 

is equal to the coefficient of x^0 ( the constant term ) in the expansion of

This product is nothing more than algebra and can be done by hand methods but it is best done by computer. If you need to I will show you how to take it over to Alpha. This is the power of generating functions, a hard problem has been reduced to a computational one. The expansion of the above product yields:

We see that the constant term is 344 and so the answer is  344 / 8^4 = 43 / 512.

Now I am not saying this is the simplest solution to this problem but these methods have the virtue of being able to solve a great many combinatorics problems with little or no change.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 4 points

It takes a little getting used to but eventually the generating functions will seem very natural.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.