• 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 no oil and the test shows that it has oil? Answer choices: 0.09 0.11 0.36 0.44
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • katieb
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
I know it's not A or C. But I tried the formula and got 1.778 but I don't know what to do from there.
  • ybarrap
Let D be the event that oil is detected Let O be the probability that oil exists We are given that \(P(O)=0.45\) and \(P(D|O)=0.8\) We also know that $$ P(D|O)=\frac{P(D\cap O)}{P(O)} $$ Where \(\bar{O}\) is the event that there is no oil Then $$ P(D\cap O)=P(D|O)P(O)=0.80\times0.45 $$ We also know that $$ P(D|O)+P(D|\bar{O})=1\\ \text{Which is the same as}\\ \frac{P(D\cap O)}{P(O)}+\frac{P(D\cap \bar{O})}{P(\bar{O})}=1\\ $$ We need \(P(D\cap \bar{O})\), the probability that oil is detected but none exists: $$ \frac{P(D\cap O)}{P(O)}+\frac{P(D\cap \bar{O})}{P(\bar{O})}=1\\ \implies P(D\cap \bar{O})=\left(1-\frac{P(D\cap O)}{P(O)}\right )P(\bar{O})\\ =(1-0.8)0.55 $$ Does this make sense?

Looking for something else?

Not the answer you are looking for? Search for more explanations.