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Statistics 8 Online
OpenStudy (anonymous):

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are below. P(A) = 0.80 P(B | A) = 0.94 P(B | A' ) = 0.79 P(C | A ∩ B) = 0.76 P(C | A ∩ B' ) = 0.56 P(C | A' ∩ B) = 0.68 P(C | A' ∩ B' ) = 0.33 (c) Compute P(B ∩ C).

OpenStudy (perl):

I can help you get started. P( B & C ) = P(B) * P( C | B )

OpenStudy (anonymous):

Thanks, but I got that. I don't know how to go from there.

OpenStudy (perl):

ok, lets use the given probabilities

OpenStudy (perl):

I would solve this by making a tree diagram

OpenStudy (perl):

|dw:1423461020448:dw|

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