Probability of a Sum Meeting a ConditionDate: 08/24/2007 at 01:01:54 From: Darren Subject: Real Numbers Two real numbers x and y are randomly chosen on a number line between 0 and 12. Find the probability that their sum is less than or equal to 5. I am not sure how to do this. My idea below is probably wrong, and I need to know how to do the problem. 0+1=1 0+2=2 0+3=3 ..................... 12+12=24 x + y <= 5 ------------- = probability # of trials Date: 08/24/2007 at 12:40:53 From: Doctor Greenie Subject: Re: Real Numbers Hi, Darren - Your thoughts about the problem only use whole numbers. The problem says you can pick ANY two real numbers between 0 and 12--like 5.189238945 and the square root of 93. So you will need a different method to attack the problem. To find an approach to the problem, let's START with your work with whole numbers and then modify that approach to include all numbers between 0 and 12. We can make a chart showing all the sums of pairs of whole numbers from 0 to 12; we get our familiar addition table, but I'm going to arrange it in an unusual manner. 0 1 2 3 4 5 6 7 8 9 10 11 12 ------------------------------------------- 12 | 12 13 14 ... | 11 | 11 12 13 ... | 10 | 10 11 12 ... | 9 | 9 10 11 ... | 8 | ... | 7 | ... | 6 | 6 7 8 ... | 5 | 5 6 7 8 ... | 4 | 4 5 6 7 8 9 10 ... | 3 | 3 4 5 6 7 8 9 10 11 12 13 14 15 | 2 | 2 3 4 5 6 7 8 9 10 11 12 13 14 | 1 | 1 2 3 4 5 6 7 8 9 10 11 12 13 | 0 | 0 1 2 3 4 5 6 7 8 9 10 11 12 | ------------------------------------------- Suppose for now that we were restricting our numbers to whole numbers, and that we wanted the sum of the two numbers to be 5 or less. Then we could get the answer directly from this table. There are 13 whole numbers from 0 to 12, so the number of sums in the table is 13*13=169. And we can count the number of sums that are 5 or less; it is 21. So using only whole numbers, the probability that our sum is 5 or less is 21/169. To modify this approach so that we consider ALL real numbers instead of just whole numbers, we can keep the same basic picture, but instead of having separate, discrete numbers horizontally and vertically, we have a continuous range of numbers. So we can think of our picture as a square 12 units wide and 12 units high: 0 1 2 3 4 5 6 7 8 9 10 11 12 12 ------------------------------------- 11 | | 10 | | 9 | | 8 | | 7 | | 6 | | 5 | | 4 | | 3 | | 2 | | 1 | | 0 ------------------------------------- The complete set of combinations of two numbers we could select from anywhere in this figure is represented by the AREA of the figure, which is 12*12 = 144. Our objective is to determine what fraction of that total area represents pairs of numbers whose sum is 5 or less. We can get an idea of how to do that by looking at the figure we had when we were using whole numbers. The pairs of numbers whose sum is exactly EQUAL to 5 lie along a diagonal line from (0,5) to (5,0). So in our figure for the case where we are allowing ANY real numbers, we can draw a boundary for the sums that are 5 or less along that diagonal line: 0 1 2 3 4 5 6 7 8 9 10 11 12 12 ------------------------------------- 11 | | 10 | | 9 | | 8 | | 7 | | 6 | | 5 \ | 4 | \ | 3 | \ | 2 | \ | 1 | \ | 0 ---------------\--------------------- The sets of combinations of two numbers we can choose whose sum is 5 or less is represented by the AREA of the triangle formed by that boundary line. That triangle is a right triangle with legs of length 5, so its area is (1/2)(5)(5) = 12.5. So the total area from which we can pick our two real numbers between 0 and 12 is 144, and the total area for which the sum of those two numbers is 5 or less is 12.5. Therefore, the probability that the two numbers we pick between 0 and 12 have a sum of 5 or less is 12.5/144 = 25/288 The method for solving the problem is the same regardless of what the maximum value of our sum is supposed to be. For the case where the sum is supposed to be at most 12, you can probably see that the "boundary line" will cut the rectangle exactly in half, so the probability will be 1/2. For the case where the sum is supposed to be (for example) at most 18, the picture is a bit different, and the calculations might be done a bit differently. Our picture would be 0 1 2 3 4 5 6 7 8 9 10 11 12 12 ------------------\------------------ 11 | \ | 10 | \ | 9 | \ | 8 | \ | 7 | \ | 6 | \ 5 | | 4 | | 3 | | 2 | | 1 | | 0 ------------------------------------- In this case, the area containing the allowable pairs of numbers is represented by the whole rectangle, MINUS the small triangular area. The area of the whole rectangle is 144; the area of the small triangle is (1/2)(6)(6) = 18. So the area of the allowable region is 144-18 = 126. And the probability that our sum is at most 18 in this case is then 126/144 = 7/8. I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ Date: 08/24/2007 at 13:58:27 From: Darren Subject: Thank you (Real Numbers) Thank you for answering my question. It is helpful to know how to do the problem. |
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