Question 47461
I'm sorry, but every time I've tried this problem, it adds up to 43 students, not 40.::
<pre><tt>
      23
     / | \
    7  7  5
   / /   \ \
  25---6---27
As you can see, this is my attempt at a triangle.
The numbers are thence explained:
the corners of the triangle, 23, 25, and 27, are respectively the number of students enrolled in <br>music, math, and comp. sci..
the 7 connecting just music and math is the 7 students in only both classes.
the 6 connecting the math and comp. sci. is the students in only the two classes.
the 5 connecting the music and comp. sci. is the only 5 doing those two classes.
and lastly the 7 connecting the 3 represents the 7 students taking all three classes.

To get the sum of all of the students taking only one course, one must take all three numbers <br>and subtract from each the numbers connecting to it.  
For example,  to get the total number of students ONLY enrolled in music, I <br>would equate 23-7-7-5.  <br>this means 23 minus the 7 math/music minus the 7 math/music/comp minus the 5 music/comp., it would give me 4.
If we do this for all of the numbers, we'd have 23-29, 25-20, and 27-18 for (respectively) <br>music, math, and comp.  The answers for each are 6, 5, and 9.
We'd be fine up to this point, saying that p(only one course) is (5+6+9)/40, or .5, until we <br>check to make sure it really is 40.  if it were, then 
<font color=green>5+6+9</font><font color=red>+5+6+7</font><font color=brown>+7</font>=<font color=orange>40</font>.
with key:
<font color=green>students enrolled in just one class</font>
<font color=red>students in just two classes</font>
<font color=brown>students in all three</font>
<font color=orange>total number of students</font>

However, this is not the case:
5+6+9+5+6+7+7=43, not 40.  So the problem is, I think, wrong.  Were it not for the bit of <br>information that the total number of students were 40, we could assume it were 43, <br>and our probability would be 30/43.  Also, suppose it were the other way around <br>and we were told there were 43 students, we could assume 3 are unenrolled in any classes.
Alas, that is as much assistance as I can give.  But ask for more if you need to!</tt></pre>