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January 24th, 2014, 07:55 AM   #1
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Another Probability Problem

One side of a triangle is 5cm long. Two numbers (don't have to be integral) are randomly selected between 0 and 10. What's the prob. that the two #s are able to be the other two sides of the triangle.
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January 24th, 2014, 08:29 AM   #2
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Re: Another Probability Problem

Let the first number be x. It's evenly distributed between 0 and 10.

The probability that the second number can form a triangle with 5 and x is:



You can work this out now using averages (given the uniform dist), rather than needing integrals of pdf's:

the average probability is 1/2 and

the average probability is 3/4.

So, overall the average probability is 3/8.
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January 27th, 2014, 04:47 PM   #3
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Re: Another Probability Problem

Hi,

What did you mean by saying "using averages"..

Thanks
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January 28th, 2014, 03:15 AM   #4
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Re: Another Probability Problem

By "using averages", I meant:

Quote:
Originally Posted by Pero
the average probability is 1/2 and

the average probability is 3/4.

So, overall the average probability is 3/8.
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January 28th, 2014, 05:56 AM   #5
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Re: Another Probability Problem

I, also, do not know what Pero means by "averaging" the probabilities. An average is itself a kind of "average" and I have never heard of averaging "averages". Here's how I would do it. In order for "x" and "y" to be lengths of two sides of a triangle with third side 5, we must have and, of course, they must be positive. On an xy-coordinate system, that gives a right triangle with legs of length 5 and so area . Since x and y, separately, are chosen, randomly (equally likely) from 0 to 10, the entire set of possibilities is a square with sides of length 10 and so area 100. "The probability that a point chosen from the larger square is in the triangle is , not .
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January 28th, 2014, 07:05 AM   #6
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Re: Another Probability Problem

First, I can't add up:

the average probability is 1/2 and

the average probability is 3/4.

So, overall the probability is (1/2)(1/2) + (1/2)(3/4) = 5/8.

That's the "average of the averages"!

I don't think the answer of 1/8 can be right, as it's too low. In any case x + y >= 5 is not the right criterion. E.g.:

x = y = 6 allows a triangle to be formed, where x = 2, y = 8 does not.

Instead the equations are:





Then, looking at the area of this in the region from 0 to 10 for x and y, gives everything except three 5x5 triangles. So, that comes to (100 - 37.5)/100 = 5/8.
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January 28th, 2014, 11:20 AM   #7
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Re: Another Probability Problem

Quote:
Originally Posted by p99410
One side of a triangle is 5cm long. Two numbers (don't have to be integral) are randomly selected between 0 and 10. What's the prob. that the two #s are able to be the other two sides of the triangle.
Have you tried with the 2 numbers being integers 1 to 9? More enjoyable :P
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January 28th, 2014, 04:27 PM   #8
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Re: Another Probability Problem

Oh, guys,
I found the answer, it's 5/8..
If you do want the answer, reply. If you still want to think more, just tell me then I'll post it. It's tricky...
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January 28th, 2014, 04:32 PM   #9
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Re: Another Probability Problem

Hi, Pero,
Quote:
for x <=5 the average probability is 1/2 and

for x >= 5the average probability is 3/4.
That is the part that I don't understand. How did you get that?

Denis, I have to say math is never enjoyable to me..
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January 29th, 2014, 12:29 AM   #10
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Re: Another Probability Problem

It's probably easier to think about areas in the region 0 <= x, y >= 10.

1/2 is just the average height of the triangle under the line from (0,0) to (5,1). And, 3/4 is the average height of the trapezium under the line from (5,1) to (10,1/2).
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