[Guest]

Dear all,

I encountered a difficult probability problem, and we know you are smart. Asking for your help to solve it.

The question is simple,

choose three length a, b, c independently from uniform [0,1] distribution. what is probability to create a triangle from a,b,c?

Thanks in advance

[curr3nt]

but it would seem that given any a or b from 0 to 1 (0 can be excluded since it wouldn't be a length?)

what is the probability that abs(a-b) < c < a+b with c not exceeding 1

[Guest]

Dear all,

I encountered a difficult probability problem, and we know you are smart. Asking for your help to solve it.

The question is simple,

choose three length a, b, c independently from uniform [0,1] distribution. what is probability to create a triangle from a,b,c?

Thanks in advance

i am assuming that what u mean by uniform [0,1] distribution is the set {0,1}, and not the actual uniform distribution which is a continuous line between 0 and one, because if it is, then there are infinite number of points between 0 and 1, and hence, infinite candidates for a, b, and c that make a triangle, and infinite candidates that dont make a triangle.

making that assumption, the only way to get a triangle is if all 3 lengths, a, b, and c get the value 1.

therefore, the probability of having a triangle is: the (probability of a getting 1) x (the probability of b getting 1) x (the probability of c getting 1)

which is: 1/2 x 1/2 x 1/2 = 0.125 or 12.5%

[Guest]

Assuming that the values for a, b, c are continuous (rather than discrete) non-negative Real Numbers, then the condition to form a triangle with possible measures a, b, c consists in:

a <> 0

b <> 0

c <> 0

That's because a triangle's side (any side) must be > 0.

Thus, the probability is as follows:

P(a<>0)*P(b<>0)*P(c<>0)

In continuous values, the probability to hit an exact number is zero.

P(x=0)=0

Therefore, the probability of hitting a number NOT zero is:

P(x<>0)=1-P(x=0)

P(x<>0)=1-0

P(x<>0)=1

Hence, the pobability of forming a triangle, given 3 random numbers (non-negative Real Numbers) is as follows:

P(a<>0)*P(b<>0)*P(c<>0)

=1*1*1

=1

Am I missing somethin'?

[Guest]

but it would seem that given any a or b from 0 to 1 (0 can be excluded since it wouldn't be a length?)

what is the probability that abs(a-b) < c < a+b with c not exceeding 1

U is uniform on [0,1]. it is continuos.

[Guest]

i am assuming that what u mean by uniform [0,1] distribution is the set {0,1}, and not the actual uniform distribution which is a continuous line between 0 and one, because if it is, then there are infinite number of points between 0 and 1, and hence, infinite candidates for a, b, and c that make a triangle, and infinite candidates that dont make a triangle.

making that assumption, the only way to get a triangle is if all 3 lengths, a, b, and c get the value 1.

therefore, the probability of having a triangle is: the (probability of a getting 1) x (the probability of b getting 1) x (the probability of c getting 1)

which is: 1/2 x 1/2 x 1/2 = 0.125 or 12.5%

I think a,b,c must match some condition to form a triangle. like a + b > c, a + c > b etc...

[Guest]

I don't know.

It need consideration.

Dear all,

I encountered a difficult probability problem, and we know you are smart. Asking for your help to solve it.

The question is simple,

choose three length a, b, c independently from uniform [0,1] distribution. what is probability to create a triangle from a,b,c?

Thanks in advance

[EventHorizon]

The answer is 1/2.

Change the problem from choosing 3 points to choosing a point in a unit cube.

This changes the problem to finding the volume of the space that satisfies the conditions that no 1 value is greater than the sum of the other two.

I drew up a couple 2d slices (keeping c a constant) to help visualize the area, and came up with an equation for the area that can produce a triangle in a given slice.

A(c=x) = 1 (the whole square)

- x²/2 (if a+b<c)

- (1-x)²/2 (if a+c<b)

- (1-x)²/2 (if b+c<a)

A(c=x) = 1 - x²/2 - (1-x)²

A(c=x) = 1 - x²/2 - 1 + 2x - x²

A(c=x) = 2x-(3/2)x²

Now I simply need to integrate from 0 to 1

The equation integrates to x²-.5x³, and plugging in 1 gives 1/2 and 0 gives 0, so the definite integral = 1/2.

[Guest]

I don't know.

It need consideration.

Thanks for your thinking. Your point is very helpful. But maybe you missed something.

1st. there is other condition needed to let a,b,c create triangle. abs(a-b) < c, abs(a-c)<b,abs(b-c) < a

2nd, you cut off corner (a+b < c) and (b+c <c). the two section intersect some part.

[EventHorizon]

Thanks for your thinking. Your point is very helpful. But maybe you missed something.

1st. there is other condition needed to let a,b,c create triangle. abs(a-b) < c, abs(a-c)<b,abs(b-c) < a

2nd, you cut off corner (a+b < c) and (b+c <c). the two section intersect some part.

1. look at abs(a-b) < c

assume that a>b.

The absolute value doesn't change anything, so it is a-b < c.

Move b to the rhs, a < b+c. I covered this case by cutting the corner piece "b+c < a".

assume that b>a.

The absolute value negates it so it is b-a < c.

Move a to the rhs, b < c+a. I covered this case by cutting the corner piece "a+c < b".

With each of the other equations you give, they also break down into one of the three corners I cut.

2. The three areas I removed correspond to when a given length is the longest (b+c<a means a is longest, etc). No two of those equations can be valid simultaneously due to there being a definite longest length if one is true.

Perhaps this picture will help

3ltri.bmp

[Guest]

1. look at abs(a-b) < c

assume that a>b.

The absolute value doesn't change anything, so it is a-b < c.

Move b to the rhs, a < b+c. I covered this case by cutting the corner piece "b+c < a".

assume that b>a.

The absolute value negates it so it is b-a < c.

Move a to the rhs, b < c+a. I covered this case by cutting the corner piece "a+c < b".

With each of the other equations you give, they also break down into one of the three corners I cut.

2. The three areas I removed correspond to when a given length is the longest (b+c<a means a is longest, etc). No two of those equations can be valid simultaneously due to there being a definite longest length if one is true.

Perhaps this picture will help

3ltri.bmp

Thanks for detailed plot.

You are right...

Thanks a lot

[EventHorizon]

Thanks for posting. It was an interesting puzzle.

(... but hopefully it wasn't homework )

Edited March 17, 2011 by EventHorizon

[Guest]

I've got 1/2.

Let's assume that a > b just for now. Given two lengths a and b, you can adjust the angle between a and b to fit the third length c. The angle is between 0, then c = a - b, and PI, then c = a + b, unless a + b > 1, in which case, c is just capped at 1 by definition (the angle can not go up to PI). This simply means that you can get the probability by integrating a, b, c over the intervals a in [0,1], b in [0,a] and c[ (a-b), max(1, a+b)].

It becomes a bit tricky (and tedious) but the max can be removed if you split the integral in 3 integrals.

The first one goes over the domain: a in [0,1/2], b in [0,a] and c[ (a-b), (a+b)] (a < 1/2 and b <a so a + b <1)

The second one goes over the domain: a in [1/2,1], b in [0,1-a] and c[ (a-b), (a+b)] (here again a + b <1), and

The third one goes over the domain: a in [1/2,1], b in [1-a,a] and c[ (a-b), 1] (here the case a +b >1 is hit)

The sum of those 3 integrals give 1/4.

Now we only considered the case a >b. The case a <b gives the same results by simple symmetry (a and b are interchangeable), so the overall probability is

1/4 + 1/ 4 = 1/2

[Guest]

3/5 ?

consider: as long as the two shortest sides total a length slightly longer than the longest side, a triangle can be made,

so the probability would just be the probability that 2 random numbers sum more than a third random number.

[Guest]

Thanks for posting. It was an interesting puzzle.

(... but hopefully it wasn't homework )

It is a interview question. hehe

[Guest]

eventhorizon solved it.