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?

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.

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.

"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.

JoeU2004 makes a good point, although I might argue his interpretation.
Is this a homework question or what? I was intentionally vague (and
leading) in my previous response.

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.

Thanks everyone for the help, and no this isnt homework, but a random
question posed by the group mentor to find out our data drilling capabilites.

Thank you & Cheers

"JoeU2004" 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.

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.

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.