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A university in Alabama has 1000 employees. Four hundred of the employees have at least 20 years of experience (event A), 100 are African American (event B), and 300 with a background in Microsoft Office 2003 (event C). Assume A,B,C are independent. What is the probability of finding an employee who meets at least two of the three criteria? Answer is suppose to be .166, but how?
 2 years ago
 2 years ago
A university in Alabama has 1000 employees. Four hundred of the employees have at least 20 years of experience (event A), 100 are African American (event B), and 300 with a background in Microsoft Office 2003 (event C). Assume A,B,C are independent. What is the probability of finding an employee who meets at least two of the three criteria? Answer is suppose to be .166, but how?
 2 years ago
 2 years ago

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frecklesBest ResponseYou've already chosen the best response.1
I thought I do P(ABC)+P(AB)+P(AC)+P(BC) =P(A)P(B)P(C)+P(A)P(B)+P(A)P(C)+P(B)P(C) but that doesn't work :(
 2 years ago

CloverpetalBest ResponseYou've already chosen the best response.0
That should work... hmmmm
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
Try drawing either a Tree Diagram or a Venn Diagram. You've counted some events (like ABC) more than once.
 2 years ago

frecklesBest ResponseYou've already chosen the best response.1
So how do I draw a venn diagram for independent events?
 2 years ago

frecklesBest ResponseYou've already chosen the best response.1
That would be three separate venn diagrams right?
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
Oops, yes, you're right. What about Tree Diagrams?
 2 years ago

frecklesBest ResponseYou've already chosen the best response.1
:( I don't know. I'm not seeing it.
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
OK. I'll draw the Tree and label it, but you fill in the probabilities (on here or on a sheet of paper). Agreed?
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
dw:1331329425257:dw
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
Sorry about the scribble on the right :)
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
Now just label the branches (ABC, ABC', AB'C, AB'C' etc etc) and fill in the probabilities. Then choose which branches to add up :)
 2 years ago

BasketWeaveBest ResponseYou've already chosen the best response.2
By branch I mean the ones on the right, by the way.
 2 years ago

frecklesBest ResponseYou've already chosen the best response.1
\[P(ABC)+P(ABC')+P(AB'C)+P(A'BC)=.012+.028+.108+.018=.166\] omg you are totally awesome
 2 years ago

frecklesBest ResponseYou've already chosen the best response.1
That is a perfect idea. Thanks.
 2 years ago
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