"Six Sigma Blackbelt" wrote:
Suppose that in a company, 29% of all vice presidents hold MBA degrees,
24% hold undergraduate business degrees, and 8% hold both.

Unfortunately, the language of those statistics is ambiguous. There are at
least two interpretations, which lead to two very different formulations.

Smartin demonstrates one interpretation, namely: the 29% who hold MBA
degrees includes the 8% that hold both degrees; likewise, the 24% who hold
BA degrees includes the 8% that hold both degrees. That certainly makes the
probability questions more interesting.

On the other hand, if I told you that a bag of marbles has 29% that are red,
24% that are white and 8% that are red and white, I doubt that you would
think the 29% red includes the 8% red and white.

So another interpretation of your situation is: 29% have only MBA degrees,
24% have only BA degrees, and 8% have both.

Failure to use the word "only" in the problem statement does not necessarily
imply that Smartin's interpretation is more correct. Arguably, if the
teacher intended Smartin's interpretation, he should have written: 29% have
at least an MBA degree, 24% have at least a BA degree, and 8% have both.

If you know which of these two interpretations is correct, I suggest that
you post a clarification in order to avoid misleading responses. Otherwise,
go back to the teacher for a clarification, then post it here.

Alternatively, it might help to know whether you are taking an introductory
class in probability/statistics or in Excel. If the former, I would guess
that Smartin's interpretation is the intended one. Otherwise, I would guess
that the teacher is trying to present a problem that has a more obvious
solution (the second interpretation), focusing on simple use of arithmetic
operators and cell references.

PS: I hope this is not a question on a final exam. Helping with homework
is one thing. Abetting cheating on an exam is something else altogether.
Fortunately, we cannot be held responsible for your malfeasance.


----- original message -----

"smartin" wrote in message
...
Six Sigma Blackbelt wrote:
Suppose that in a company, 29% of all vice presidents hold MBA degrees,
24% hold undergraduate business degrees, and 8% hold both. A vice
president is to be selected randomly.

a) What is the probability that the vice president holds either an MBA or
an undergraduate business degree (or both)
b) What is the probability that the vice president holds neither?


To understand the probabilities try making a little table and fill in the
blanks...

MBA no MBA |
UB 8 ? | 24
no UB ? ? | ?
----------------------+-----
29 ? | 100

Then the formulas should become obvious.