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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: $For \ x \le 5 \ p= \frac{2x}{10} \ and \ for \ x \ge 5 \ p = \frac{15-x}{10}$ You can work this out now using averages (given the uniform dist), rather than needing integrals of pdf's: $For \ x \le 5$ the average probability is 1/2 and $For \ x \ge 5$ 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
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Re: Another Probability Problem

By "using averages", I meant:

Quote:
 Originally Posted by Pero $For \ x \le 5$ the average probability is 1/2 and $For \ x \ge 5$ the average probability is 3/4. So, overall the average probability is 3/8.

 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 $x+ y\ge 5$ 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 $\frac{25}{2}$. 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 $\frac{\frac{25}{2}}{100}= \frac{25}{200}= \frac{1}{8}$, not $\frac{3}{8}$.
 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: $For \ x \le 5$ the average probability is 1/2 and $For \ x \ge 5$ 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: $x \le 5: \ 5-x \le y \le 5+x$ $x \ge 5: \ y \ge x-5$ 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
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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

 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
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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..

 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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