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January 24th, 2014, 07:55 AM  #1 
Member Joined: Dec 2013 Posts: 31 Thanks: 0  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.

January 24th, 2014, 08:29 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  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. 
January 27th, 2014, 04:47 PM  #3 
Member Joined: Dec 2013 Posts: 31 Thanks: 0  Re: Another Probability Problem
Hi, What did you mean by saying "using averages".. Thanks 
January 28th, 2014, 03:15 AM  #4  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Another Probability Problem
By "using averages", I meant: Quote:
 
January 28th, 2014, 05:56 AM  #5 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  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 xycoordinate 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 .

January 28th, 2014, 07:05 AM  #6 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  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. 
January 28th, 2014, 11:20 AM  #7  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,465 Thanks: 949  Re: Another Probability Problem Quote:
 
January 28th, 2014, 04:27 PM  #8 
Member Joined: Dec 2013 Posts: 31 Thanks: 0  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... 
January 28th, 2014, 04:32 PM  #9  
Member Joined: Dec 2013 Posts: 31 Thanks: 0  Re: Another Probability Problem
Hi, Pero, Quote:
Denis, I have to say math is never enjoyable to me..  
January 29th, 2014, 12:29 AM  #10 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  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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