Alright, seeing your other response, I can at least walk you through the
diagram.

We know each VP falls into one of four categories: Has MBA only, Has UB
only, Has both, Has neither. These possibilities are represented by the
four cells in the upper left area of the diagram.

You gave us that 8% have both, so I placed 8 at the intersection of MBA
and UB.

29% have MBA, but we don't know if they have UB or not. But we do know
all the MBA holders make up 29% of the VPs, so 29 goes in the subtotal
cell for MBA. Can you guess, at this point, what percentage has MBA but
no UB? Yes-- it's 21%

You can also immediately find out what percentage has No MBA. Since 29%
have MBA, the remaining group must be 71%, since these two populations
must make up 100% of the VPs.

If you followed me this far you should have no trouble filling in the rest.

Then to the specific questions...

a) Is already given. 29 have MBA +24 have UB. Some of these have both,
some only one or the other.

b) Must be derived by filling in the table. The intersection of "No MBA"
and "No UB" is what we want here.

Hope this helps!



Six Sigma Blackbelt wrote:
smartin

Thanks for the lead, but I need assistance figuring out the formula to solve
the word problem..

Cheers & thanks

"smartin" wrote:

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.