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freckles
 4 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?
freckles
 4 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?

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freckles
 4 years ago
Best ResponseYou've already chosen the best response.1I 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 :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That should work... hmmmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try drawing either a Tree Diagram or a Venn Diagram. You've counted some events (like ABC) more than once.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops, yes, you're right. What about Tree Diagrams?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK. I'll draw the Tree and label it, but you fill in the probabilities (on here or on a sheet of paper). Agreed?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1331329425257:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry about the scribble on the right :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now just label the branches (ABC, ABC', AB'C, AB'C' etc etc) and fill in the probabilities. Then choose which branches to add up :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By branch I mean the ones on the right, by the way.

freckles
 4 years ago
Best 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

freckles
 4 years ago
Best ResponseYou've already chosen the best response.1That is a perfect idea. Thanks.
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