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Algebra 8 Online
OpenStudy (anonymous):

Jason inherited a piece of land from his great-uncle. Owners in the area claim that there is a 45% chance that the land has oil. Jason decides to test the land for oil. He buys a kit that claims to have an 80% accuracy rate of indicating oil in the soil. What is the probability that the land has oil and the test predicts it? how do you solve this?

OpenStudy (anonymous):

Denote \(A\) to be the event that oil is present, so \(P(A)=0.45\). Then \(P(A')\), the complement of \(A\), is 0.55. Denote \(K\) to be the event that oil is detected by the kit, so \(P(K)=0.80\) and \(P(K')=0.20\). You are asked to find \(P(A\cap K)\). Use the conditional probability formula, \[P(A|K)=\frac{P(A\cap K)}{P(K)},~~P(K|A)=\frac{P(A\cap K)}{P(A)}\] Of course, you have to determine the conditional probabilities first.

OpenStudy (anonymous):

@SithsAndGiggles so what would be the answer here

OpenStudy (anonymous):

i suck at putting formulas in calculator

OpenStudy (anonymous):

0.09 0.11 0.36 0.44

OpenStudy (anonymous):

I must have misspoke/-typed. The test for oil is 80% accurate, meaning that the test is positive 80% of the time if oil is actually present, so \(P(K|A)=0.80\). Then \[P(K|A)=\frac{P(A\cap K)}{P(A)}~~\iff~~0.80=\frac{P(A\cap K)}{0.45}~~\iff~~P(A\cap K)=0.36\]

OpenStudy (anonymous):

thank you thank you well helped

OpenStudy (anonymous):

yw

OpenStudy (anonymous):

@jcr268 Wait what was the answer? I don't understand that formula or how to calculate it.

OpenStudy (anonymous):

http://openstudy.com/users/jcr268 help plz

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