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### Probability of a Sum Meeting a Condition

```Date: 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 -

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.

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

- 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.
```
Associated Topics:
High School Linear Equations
High School Probability

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