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It's a trick question. The answer has already been given to you.
 
Upon further review, you are correct.

The answer would be 50+41-19=72, or 72 in 100.

Nice catch.

jonnythan:
The 19% isn't meaningless. The 19% indicates that 19 of your 91 people are duplicates. There's no way to have 50 people who work, 41 who go to school, 19 who do both, and only 9 who do neither.
 
Don Burke:
I never said it was meaningless.

I said it was meaningless for this purpose, which is true.
No, it's not meaningless. The 19% tells you that 19 people belong to both the set of workers and the set of students. In other words, 91% is wrong because you didn't subtract the 19 from it.
 
Don Burke:
I never said it was meaningless.

I said it was meaningless for this purpose, which is true.
Why? I find Jonnythan's analysis to be clear, simple, and accurate.

The statements of 50% work, 41% school, 19% both can also be expressed as 31% work only, 22% school only, 19% both (See Jonnythan's diagram).
 
Charlie99:
Why? I find Jonnythan's analysis to be clear, simple, and accurate.
My bust. I made the mistake of answering before my tenth cup of coffee.
 
Jonnythan's calculation is quite correct as the problem is stipulated.
Out of a universe of 100
50 work
19 do both which is the overlap
leaves 22 that only go to school
so 72 (=72%) do something
leaving 28 (=28%) lazy bums.

So:
What is the probability that by randomly choosing someone in this age group that you will pick someone who is working full-time Answer=50%
or going to school Answer=41%
or both? Answer = 19%

Just for the record
or either? Answer=72%
 
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